1. Tardigrade
  2. Question
  3. Chemistry
  4. Consider the following four electrodes: P=Cu2+(0.0001 M) / Cu(s) Q=Cu2+(0.1 M)/ Cu(s) R=Cu2+(0.01 M)/ Cu(s) S=Cu2+ (0.001 M)/ Cu(s) If the standard reduction potential of Cu2+/Cu is +0.34 V, the reduction potentials in volts of the above electrodes follow the order

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

KEAMKEAM 2010Electrochemistry

Solution:

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