Question

Consider the following four electrodes :
$P=C{u}^{2+}\left(0.0001M\right)/C{u}_{\left(s\right)}$
$Q=C{u}^{2+}\left(0.1M\right)/C{u}_{\left(s\right)}$
$R=C{u}^{2+}\left(0.01M\right)/C{u}_{\left(s\right)}$
$S=C{u}^{2+}\left(0.001M\right)/C{u}_{\left(s\right)}$
If the standard reduction potential of $C{u}^{2+}/Cu$ is +0.34 V, the reduction potentials in volts of the above electrodes follow the order

• A. $P>S>R>Q$
• B. $S>R>Q>P$
• C. $R>S>Q>P$
• D. $Q>R>S>P$
• E. $P>Q>R>S$

Answer 1

Answer: (D)

$E=E^{\circ }+\frac{0.591}{n}log\left[M^{n+}\right]$
Lower the concentration of $M^{n+},$ lower is the E.