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Q.(a)(i) [2 marks] Can we store copper sulphate in a vessel made of iron? Justify your answer. (a)(ii) [3 marks] Calculate the resistance of 0.01N solution of an electrolyte whose equivalent conductivity is 420 ohm^-1 cm^2 equiv^-1. (The cell constant of the cell is 0.88 cm^-1) OR (b)(i) [2 marks] Why is the alternating current used for measuring the resistance of an electrolytic solution? (b)(ii) [3 marks] Calculate the e.m.f of the cell in which the following reaction takes place: Ni(s) + 2Ag+(0.002M) → Ni2+(0.160M) + 2Ag(s). Given that the standard electrode potential of the cell is 1.05 V.

Mizoram MbseMizoram Board of School Education HSSLC 2021Subjective· 5mImportance★★★★★
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(a)(i) Fe displaces Cu from CuSO4CuSO_4 (Fe is above Cu in the reactivity/electrochemical series). (a)(ii) R=G∗/κR = G^*/\kappa, with κ\kappa from Λeq\Lambda_{eq} and normality. / (b)(i) AC prevents electrolysis of the solution. (b)(ii) Use the Nernst equation.

(a)(i) Can CuSO4CuSO_4 be stored in an iron vessel?

No. Iron (Fe) is more reactive than copper (Cu) — it lies above copper in the reactivity/electrochemical series (i.e. Fe has a more negative standard reduction potential than Cu). A more reactive metal displaces a less reactive metal from a solution of its salt:

Fe(s)+CuSO4(aq)→FeSO4(aq)+Cu(s)Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s)

This reaction would slowly dissolve/corrode the iron vessel while depositing copper on it, so copper sulphate solution cannot be safely stored in an iron container.

(a)(ii) Resistance of the 0.01 N solution:

Given: equivalent conductivity Λeq=420 Ω−1cm2 equiv−1\Lambda_{eq} = 420\ \Omega^{-1}cm^2\,equiv^{-1}; normality N=0.01N = 0.01; cell constant G∗=0.88 cm−1G^* = 0.88\ cm^{-1}.

Specific conductance: κ=Λeq×N1000=420×0.011000=4.2×10−3 Ω−1cm−1\kappa = \dfrac{\Lambda_{eq} \times N}{1000} = \dfrac{420 \times 0.01}{1000} = 4.2\times10^{-3}\ \Omega^{-1}cm^{-1}.

Since κ=G∗R\kappa = \dfrac{G^*}{R}:

R=G∗κ=0.884.2×10−3≈209.5 ΩR = \dfrac{G^*}{\kappa} = \dfrac{0.88}{4.2\times10^{-3}} \approx 209.5\ \Omega

(OR) (b)(i) Why AC, not DC, for measuring electrolytic resistance?

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