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Consider the following four electrodes: A = Cu²⁺(0.0001 M)/Cu(s); B = Cu²⁺(0.1 M)/Cu(s) ; C = Cu²⁺(0.01 M)/Cu(s); D = Cu²⁺(0.001 M)/Cu(s). If the standard reduction potential of Cu⁺² /Cu is +0.34V, the reduction potentials (in volts) of the above electrodes follow the order

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$A > D > C > B$

$B > C > D > A$

$C > D > B > A$

$A > B > C > D$

answer is B.

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Detailed Solution

${Cu}^{+ 2} + 2 e^{-} \to Cu$

$E_{RP} = E_{RP}^{0} + \frac{0.0591}{2} \log [{Cu}^{+ 2}]$

$E_{RP} \infty {[M^{+ n}]}_{cathode}$

$R.P. = A < D < C < B$

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