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If the $0.05$ molar solution of $M^{+}$ is replaced by a $0.0025 M$ molar $M^{+}$ solution, then the magnitude of cell potential would be: $(E_{cell (1)} = 70 mV)$

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$35 mV$

$70 mV$

$140 mV$

$700 mV$

answer is C.

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Detailed Solution

$E_{cell}^{o} = 0, for every concentration cell E = 0 - \frac{0.059}{1} \log \frac{{[M^{-}]}_{anode}}{{[M^{+}]}_{cathode}} = \frac{- 0.059}{1} \log [0.0025] = + 153 mV$
It is close to 140 mV.

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