Q18 · CSIR NET Chemistry, June 2012
Paper: CSIR NET June 2012 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium
The correct value of $\mathrm{E}^{\circ}$, of a half cell in the following graph of E vs $\log \mathrm{m}$ (molality) is: 
$\log \mathrm{m}$
$\log \mathrm{m}$(a)CC'/ AC'
(b)AB'
(c)BB'
(d)CC'
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
E = E° + (RT/nF) ln m, so E = E° where log m = 0. That is the intercept on the E axis, BB′ (c).
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