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Q.

The electrode potential of $M^{+ n} + {ne}^{-} \to M$ Changes by 0.03V for a tenfold change in concentration of $M^{+ n}$ , what is the value of n?

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a

2

b

1

c

3

d

4

answer is A.

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Detailed Solution

$E_{1} = E ° - \frac{0.06}{n} \log \frac{1}{C_{1}}$

$E_{2} = E ° - \frac{0.06}{n} \log \frac{1}{C_{2}}$

$E_{2} - E_{1} = 0.03 = \frac{- 0.06}{n} \frac{(- 1)}{1}$

$n = 2$

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The electrode potential of M+n+ne-→M Changes by 0.03V for a tenfold change in concentration of M+n, what is the value of n?