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Q.

The solution of nickel sulphate in which nickel rod is dipped is diluted 10 times. The potential of nickel

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a

decreases by 60 mV

b

increases by 30V

c

decreases by 30 mV

d

decreases by 60V

answer is C.

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Detailed Solution

Half cell reaction is Ni⁺² +2e^- → Ni

Applying Nernst equation

{E_{N{i^{ + 2}}/Ni}} = E_{N{i^{ + 2}}/Ni}^o - frac{{0.0591}}{2}log frac{{left[ {Ni} right]}}{{left[ {N{i^{ + 2}}} right]}}

{E_{N{i^{ + 2}}/Ni}} = E_{N{i^{ + 2}}/Ni}^o + frac{{0.0591}}{2}log left[ {N{i^{ + 2}}} right]

Since the solutioin is 10 times diluted; [Ni]⁺² decreases 10 times. [Ni]⁺² =0.1M

{E_{N{i^{ + 2}}/Ni}} = E_{N{i^{ + 2}}/Ni}^o + frac{{0.0591}}{2}log {10^{ - 1}} = E_{N{i^{ + 2}}/Ni}^o - 0.0295

Reduction potential decreases by approximately 30mV

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The solution of nickel sulphate in which nickel rod is dipped is diluted 10 times. The potential of nickel