Q.Calculate the emf of the cell in which the following reaction takes place:
$Ni(s) + 2Ag^+(0.002\ M) \rightarrow Ni^{2+}(0.160\ M) + 2Ag(s)$
Given that $E^\circ_{cell} = 1.05\ V$.

Concept: Cell Representation & Nernst Equation

The cell is a concentration-type cell where the emf depends on the ion concentrations via the Nernst equation.

Step 1 – Write the Nernst equation for the cell

For the reaction:

$Ni(s) + 2Ag^+(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)$

Number of electrons transferred, $n = 2$.

The Nernst equation at 298 K is:

Step 2 – Substitute the given values

$E^\circ_{cell} = 1.05\ \text{V}$, $[\text{Ni}^{2+}] = 0.160\ \text{M}$, $[\text{Ag}^+] = 0.002\ \text{M}$

Step 3 – Simplify the log term …

This is a classic Nernst equation problem. The key idea is that the standard cell potential ($E^\circ_{cell}$) is measured when all ionic concentrations are 1 M. Here, the concentrations are different, so the actual emf shifts according to the reaction quotient.

The Nernst equation for a cell reaction at 298 K (room temperature, assumed unless stated otherwise) is:

where $n$ is the number of electrons transferred in the balanced reaction, and $Q$ is the reaction quotient.

Let’s work through it step by step.

1. Identify $n$, the number of electrons transferred.

   The reaction is:

   $Ni(s) + 2Ag^+(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)$

   Nickel goes from 0 to +2 (loses 2 electrons). Each silver ion goes from +1 to 0 (gains 1 electron), and there are two silver ions — so total electrons gained = 2.

   Therefore, $n = 2$.

2. Write the reaction quotient $Q$.

   For the reaction:

Solids (Ni and Ag) do not appear in $Q$ because their activities are 1.

Given: $[Ni^{2+}] = 0.160\ M$, $[Ag^+] = 0.002\ M$.

So:

3. Apply the Nernst equation.

First, compute $\log(40,000)$. Since $40,000 = 4 \times 10^4$,

(You can also do $\log(4 \times 10^4) = \log 4 + 4$ directly.) …

Method: Nernst Equation for Cell EMF Under Non-Standard Conditions

The Nernst equation allows us to calculate the cell potential when concentrations are not at standard conditions (1 M).

Step 1: Write the Nernst Equation

For a general cell reaction:

The Nernst equation at 298 K is:

Where:

• $n$ = number of electrons transferred
• $Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$

Step 2: Identify $n$ from the reaction

Given:

• Ni loses 2 electrons: $Ni \rightarrow Ni^{2+} + 2e^-$
• Each Ag⁺ gains 1 electron: $2Ag^+ + 2e^- \rightarrow 2Ag$

$n = 2$

Step 3: Write the reaction quotient $Q$

Solids (Ni and Ag) are not included in $Q$.

Substitute the given concentrations:

Step 4: Apply the Nernst equation

Given $E^\circ_{cell} = 1.05\ V$: …

Common Mistakes: EMF of the Ni | Ni²⁺ || Ag⁺ | Ag Cell (Nernst Equation)

✗ Mistake 1: Getting the Number of Electrons Transferred (n) Wrong

The error: Students see "2Ag⁺" in the equation and set $n=1$ (copying the coefficient of a single Ag⁺), or they double-count and use $n=4$.

Why it's wrong: $n$ is the total number of electrons transferred in the balanced overall reaction. Here, Ni loses 2 electrons ($Ni \rightarrow Ni^{2+} + 2e^-$) and each of the 2 Ag⁺ ions gains 1 electron ($2Ag^+ + 2e^- \rightarrow 2Ag$) — so exactly 2 electrons are transferred overall, giving $n = 2$.

How to avoid: Always write both half-reactions, balance the electrons between them, and read off $n$ from the balanced overall equation — never just copy a stoichiometric coefficient.

✗ Mistake 2: Writing the Reaction Quotient Q Incorrectly

The error: Students write $Q = \frac{[Ag^+]^2}{[Ni^{2+}]}$ (inverted), or forget to square $[Ag^+]$.

Why it's wrong: For $Ni(s) + 2Ag^+(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)$, $Q$ is products over reactants, each raised to its stoichiometric coefficient, with solids omitted: $Q = \dfrac{[Ni^{2+}]}{[Ag^+]^2}$.

How to avoid: Write the balanced equation first, then build $Q$ directly from it — products (aqueous species) on top, reactants (aqueous species) on the bottom, solids excluded.

✗ Mistake 3: Forgetting the Minus Sign in the Nernst Equation

The error: Students write $E_{cell} = E^\circ_{cell} + \frac{0.0591}{n}\log Q$.

Why it's wrong: The correct form at 298 K is $E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q$. Here $Q = \frac{0.160}{(0.002)^2} = 40{,}000 \gg 1$, so the correction term is subtracted, pulling $E_{cell}$ below $E^\circ_{cell} = 1.05\ V$.

How to avoid: Memorise the formula with the minus sign, and sanity-check: if $Q>1$ (reaction has moved toward products relative to standard conditions), $E_{cell}$ should be less than $E^\circ_{cell}$.

✗ Mistake 4: Arithmetic Errors in the Logarithm …