Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s

Exercises · 2.4

Q. Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

(i)

2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd

(ii)

Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)

Calculate the

\Delta_rG^\circ

and equilibrium constant of the reactions.

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Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is given by:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
$R$ = universal gas constant (8.314 J/mol·K)
$T$ = temperature in Kelvin
$n$ = number of moles of electrons transferred in the balanced half-reactions
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

Solids (Zn, Cu) do not appear in $Q$ because their concentrations are constant (activity = 1).

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V.

$E = 1.10 - \frac{0.0591}{2} \log_{10} \frac{0.1}{1.0}$ …

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

Why does a cell's voltage change when concentrations change? The Nernst equation answers this — but the reason lies in the link between chemical free energy and electrical work.

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

A galvanic cell does electrical work. The maximum useful work a cell can do equals the change in Gibbs free energy ( $\Delta G$ ):

\Delta G = -nFE_{\text{cell}}

Where:

$n$ = moles of electrons transferred
$F$ = Faraday constant ( $96,485 \, \text{C mol}^{-1}$ )
$E_{\text{cell}}$ = cell potential (volts)

Why negative? A spontaneous reaction has $\Delta G < 0$ and $E_{\text{cell}} > 0$ — the negative sign makes this consistent.

2. The Chemical Side: $\Delta G$ Depends on Concentration

For a general redox reaction:

aA + bB \rightarrow cC + dD

The Gibbs free energy under non-standard conditions is:

\Delta G = \Delta G^\circ + RT \ln Q

Where $Q$ is the reaction quotient:

Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Why this form? It comes from the relationship between chemical potential and concentration — the entropy of mixing drives concentration dependence.

3. Combining Both Sides: The Derivation

Set the electrical work equal to the chemical free energy change:

-nFE_{\text{cell}} = -nFE^\circ_{\text{cell}} + RT \ln Q

Divide both sides by $-nF$ :

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q

This is the Nernst equation.

4. The "Why" in Plain Terms

Concept	Physical Meaning
$E^\circ_{\text{cell}}$	Voltage when all species are at 1 M (standard state)
$-\frac{RT}{nF} \ln Q$	Correction factor — adjusts voltage for real concentrations
$Q$	Tells you how far the reaction is from equilibrium

Key insight: When $Q = K$ (equilibrium), $E_{\text{cell}} = 0$ — the battery is dead because no net reaction occurs.

5. The Common Form (log base 10)

At $25^\circ \text{C}$ ( $298 \, \text{K}$ ):

\frac{RT}{F} \ln 10 \approx 0.0592 \, \text{V}

So:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log_{10} Q

Why convert to log? Exam convenience — most concentration values are powers of 10.

--- …

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The standard cell potential is the difference between the standard reduction potentials of the cathode and anode. For each reaction, we identify the half-reactions, look up their standard potentials, and then use $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ . From $E^\circ_{\text{cell}}$ , we calculate $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$ and the equilibrium constant $K$ using $\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}$ .

The Core Idea: Why Cell Potential Tells Us About Spontaneity

A galvanic cell works because electrons flow spontaneously from a stronger reducing agent (anode, where oxidation happens) to a weaker one (cathode, where reduction happens). The driving force is the difference in their tendencies to gain electrons — measured as the standard reduction potential, $E^\circ$ .

The Nernst equation at standard conditions gives us the cell potential directly:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

A positive $E^\circ_{\text{cell}}$ means the reaction is spontaneous. From there, the Gibbs free energy change tells us the maximum useful work obtainable:

\Delta_r G^\circ = -nFE^\circ_{\text{cell}}

And the equilibrium constant $K$ tells us how far the reaction goes:

\Delta_r G^\circ = -RT \ln K \quad \Rightarrow \quad \ln K = \frac{nFE^\circ_{\text{cell}}}{RT}

At 298 K, using $F = 96485\ \text{C mol}^{-1}$ and $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ , we often use the convenient form:

\log_{10} K = \frac{nE^\circ_{\text{cell}}}{0.059}

Let's apply this to each reaction.

Reaction (i): $2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)$

1. Identify the half-reactions

Oxidation (anode): Cr metal loses electrons to become Cr³⁺.

$Cr(s) \rightarrow Cr^{3+}(aq) + 3e^-$

Standard reduction potential (for the reverse): $E^\circ_{Cr^{3+}/Cr} = -0.74\ \text{V}$
Reduction (cathode): Cd²⁺ gains electrons to become Cd metal.

$Cd^{2+}(aq) + 2e^- \rightarrow Cd(s)$

Standard reduction potential: $E^\circ_{Cd^{2+}/Cd} = -0.40\ \text{V}$

Watch out

A common mistake is to use the oxidation potential directly. Always use reduction potentials from the table, and subtract the anode's reduction potential from the cathode's.

2. Calculate $E^\circ_{\text{cell}}$

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.40) - (-0.74) = +0.34\ \text{V}

The positive value confirms the reaction is spontaneous as written.

3. Determine $n$ , the number of electrons transferred

Look at the balanced equation: 2 Cr atoms each lose 3 electrons → total 6 electrons lost. 3 Cd²⁺ ions each gain 2 electrons → total 6 electrons gained. So $n = 6$ .

4. Calculate $\Delta_r G^\circ$

\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -6 \times 96485 \times 0.34

\Delta_r G^\circ = -196,829.4\ \text{J mol}^{-1} \approx -196.8\ \text{kJ mol}^{-1}

The negative sign means the reaction is spontaneous and can do useful work.

5. Calculate the equilibrium constant $K$

Using $\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}$ :

\ln K = \frac{6 \times 96485 \times 0.34}{8.314 \times 298}

\ln K = \frac{196829.4}{2477.572} \approx 79.45

So $K = e^{79.45}$ . That's an astronomically large number — the reaction goes essentially to completion.

Using the base-10 shortcut at 298 K:

\log_{10} K = \frac{nE^\circ_{\text{cell}}}{0.059} = \frac{6 \times 0.34}{0.059} = \frac{2.04}{0.059} \approx 34.58

So $K \approx 10^{34.58}$ , consistent with the above.

Tip

When $E^\circ_{\text{cell}}$ is positive and $n$ is large, $K$ becomes enormous — the reaction is product-favoured overwhelmingly.

Reaction (ii): $Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$

1. Identify the half-reactions

Oxidation (anode): Fe²⁺ loses an electron to become Fe³⁺. …

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Method: Standard Cell Potential from Standard Reduction Potentials

We use the Standard Reduction Potential Table and the Nernst Equation (in its standard form) to find $E^\circ_{\text{cell}}$ , then calculate $\Delta_r G^\circ$ and $K$ .

Step 1 — Write half-reactions and look up $E^\circ$ values

From the standard reduction potential table (at 298 K):

(i)

$Cd^{2+}(aq) + 2e^- \rightarrow Cd(s)$ $E^\circ = -0.40 \, \text{V}$
$Cr^{3+}(aq) + 3e^- \rightarrow Cr(s)$ $E^\circ = -0.74 \, \text{V}$

(ii)

$Ag^+(aq) + e^- \rightarrow Ag(s)$ $E^\circ = +0.80 \, \text{V}$
$Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)$ $E^\circ = +0.77 \, \text{V}$

Step 2 — Identify cathode (reduction) and anode (oxidation)

Rule: The half-reaction with the higher (more positive) reduction potential undergoes reduction at the cathode.

(i)

$Cd^{2+}/Cd$ : $E^\circ = -0.40 \, \text{V}$ (higher) → Cathode (reduction)
$Cr^{3+}/Cr$ : $E^\circ = -0.74 \, \text{V}$ (lower) → Anode (oxidation)

(ii)

$Ag^+/Ag$ : $E^\circ = +0.80 \, \text{V}$ (higher) → Cathode
$Fe^{3+}/Fe^{2+}$ : $E^\circ = +0.77 \, \text{V}$ (lower) → Anode

Step 3 — Calculate $E^\circ_{\text{cell}}$

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

(i)

E^\circ_{\text{cell}} = (-0.40) - (-0.74) = +0.34 \, \text{V}

(ii)

E^\circ_{\text{cell}} = (+0.80) - (+0.77) = +0.03 \, \text{V}

Both are positive, confirming spontaneous reactions.

Step 4 — Calculate $\Delta_r G^\circ$

\Delta_r G^\circ = -n F E^\circ_{\text{cell}}

Where:

$n$ = number of moles of electrons transferred (from balanced equation)
$F = 96485 \, \text{C mol}^{-1}$ (Faraday constant)

(i)

Balancing: $2Cr \rightarrow 2Cr^{3+}$ loses $6e^-$ ; $3Cd^{2+} \rightarrow 3Cd$ gains $6e^-$

So $n = 6$

\Delta_r G^\circ = -6 \times 96485 \times 0.34 = -1.97 \times 10^5 \, \text{J mol}^{-1}

\boxed{\Delta_r G^\circ = -197 \, \text{kJ mol}^{-1}}

(ii)

$Fe^{2+} \rightarrow Fe^{3+}$ loses $1e^-$ ; $Ag^+ \rightarrow Ag$ gains $1e^-$

So $n = 1$

$\Delta_r G^\circ = -1 \times 96485 \times 0.03 = -2.89 \times 10^3 \, \text{J mol}^{-1}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Writing the wrong cell representation

Students often write the cell diagram in the wrong order or forget the phase boundaries.

How to avoid:

Remember the mnemonic: Anode on Left, Cathode on Right → Always Leave Cathode Right.
Write the anode (oxidation) first, then the cathode (reduction).
Use single vertical lines | for phase boundaries and double lines || for the salt bridge.

Correct representation for (i):

Cr(s) | Cr^{3+}(aq) || Cd^{2+}(aq) | Cd(s)

Correct representation for (ii):

Pt(s) | Fe^{2+}(aq), Fe^{3+}(aq) || Ag^+(aq) | Ag(s)

Note: For (ii), both $Fe^{2+}$ and $Fe^{3+}$ are in solution, so we use an inert electrode (Pt). Write both ions separated by a comma.

Mistake 2: Using wrong $E^\circ$ values or wrong sign convention

Students often pick the wrong half-cell potential from the table or forget to reverse the sign for oxidation.

How to avoid:

Always use standard reduction potentials from the table.
For the anode (oxidation), reverse the sign of the given reduction potential.
Then compute:

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Example for (i):

$Cr^{3+}/Cr$ : $E^\circ = -0.74\ \text{V}$ (reduction)
$Cd^{2+}/Cd$ : $E^\circ = -0.40\ \text{V}$ (reduction)

Correct calculation:

Cathode (reduction): $Cd^{2+} + 2e^- \rightarrow Cd$ , $E^\circ = -0.40\ \text{V}$
Anode (oxidation): $Cr \rightarrow Cr^{3+} + 3e^-$ , $E^\circ = +0.74\ \text{V}$ (sign reversed)

E^\circ_{cell} = (-0.40) - (-0.74) = +0.34\ \text{V}

Mistake 3: Forgetting to balance electrons before using Nernst equation

The Nernst equation requires the balanced number of electrons ( $n$ ). Students often use $n$ from an unbalanced half-reaction.

How to avoid:

Balance the overall reaction first.
Find the LCM of electrons transferred in oxidation and reduction.

For (i):

Oxidation: $Cr \rightarrow Cr^{3+} + 3e^-$ (×2)
Reduction: $Cd^{2+} + 2e^- \rightarrow Cd$ (×3)
Total electrons transferred: $n = 6$

For (ii):

Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
Reduction: $Ag^+ + e^- \rightarrow Ag$
Total electrons transferred: $n = 1$

Mistake 4: Using wrong formula for $\Delta_r G^\circ$ and $K$

Students often confuse the sign or units.

How to avoid:

Use the correct formula:

\Delta_r G^\circ = -n F E^\circ_{cell}

$F = 96485\ \text{C mol}^{-1}$ (Faraday constant)
$\Delta_r G^\circ$ comes out in J/mol — convert to kJ/mol by dividing by 1000.

For equilibrium constant $K$ :

\Delta_r G^\circ = -RT \ln K

$\ln K = \frac{n F E^\circ_{cell}}{RT}$ …

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←Previous question2.3 — Depict the galvanic cell in which the reaction Zn(s) + 2Ag^+(aq) arrow…→Next question2.5 — Write the Nernst equation and emf of the following cells at 298 K: (i)…

Q.Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

The Core Idea: Why Cell Potential Tells Us About Spontaneity

Reaction (i): 2Cr(s)+3Cd2+(aq)→2Cr3+(aq)+3Cd(s)2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)2Cr(s)+3Cd2+(aq)→2Cr3+(aq)+3Cd(s)

1. Identify the half-reactions

2. Calculate Ecell∘E^\circ_{\text{cell}}Ecell∘​

3. Determine nnn, the number of electrons transferred

4. Calculate ΔrG∘\Delta_r G^\circΔr​G∘

5. Calculate the equilibrium constant KKK

Reaction (ii): Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)

1. Identify the half-reactions

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Reaction (i): $2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)$

2. Calculate $E^\circ_{\text{cell}}$

3. Determine $n$ , the number of electrons transferred

4. Calculate $\Delta_r G^\circ$

5. Calculate the equilibrium constant $K$

Reaction (ii): $Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$

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