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  4. Cu +2+2 e - arrow Cu ; log [ Cu +2] vs E RP graph is of the type as shown in figure where OA =0.34 volt. Then electrode potential of the half cell of Cu / Cu +2(0.1 M ) will be:-

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Q. $Cu ^{+2}+2 e ^{-} \rightarrow Cu ; \log \left[ Cu ^{+2}\right]$ vs $E _{ RP }$ graph is of the type as shown in figure where $OA =0.34$ volt. Then electrode potential of the half cell of $Cu / Cu ^{+2}(0.1\, M )$ will be:-Chemistry Question Image

Solution:

$E _{ RP }= E _{ RP }^{ o }-\frac{0.0591}{ n } \log _{10} \frac{1}{\left[ Cu ^{+2}\right]}$
$E _{ RP }= E _{ RP }^{ o }+\frac{0.0591}{ n } \log _{10}\left[ Cu ^{+2}\right]$
$y = c + m x$
$E _{ RP }^{ o }=0.34\, V$
$E _{ OP }^{ o }=-0.34\, V$
$E _{ OP }= E _{ OP }^{ o }-\frac{0.0591}{2} \log _{10}\left[ Cu ^{+2}\right]$