CBSE Class 12 Chemistry 2015 Solved Paper

Question : 21

Total: 26

Calculate emf of the following cell at 25∘C :
‌Fe‌|Fe2+(0.001M)∥H+(0.01M)| H2(g)(1‌ bar ‌)∣Pt( s)
‌E∘(Fe2+∣Fe)=−0.44VE∘ (H+∣H2)=0.00V

Solution:

The cell reaction:
‌Fe( s)+2H+(aq) ────▸Fe2+(aq) +H2(g)
‌E°‌cell ‌=Ec∘−Ea∘
‌‌=[0−(−0.44)]V=0.44V
‌E‌cell ‌=E°‌cell ‌−‌

0.059

‌log‌

[Fe2+]

[H+]2

‌E‌cell ‌=0.44V−‌

0.059

‌log‌

(0.001)

(0.01)2

‌‌=0.44V−‌

0.059

‌log(10)
‌=0.44V−0.0295V
‌‌≈0.410V

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