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:-

Question

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:- (A) (-0.34+(0.0591/2)) volt (B) (0.34+0.0591) volt (C) 0.34 volt (D) None of these

Tardigrade Admin · Accepted Answer

E RP = E RP o -(0.0591/ n ) log 10 (1/[ Cu +2]) E RP = E RP o +(0.0591/ n ) log 10[ Cu +2] y = c + m x E RP o =0.34 V E OP o =-0.34 V E OP = E OP o -(0.0591/2) log 10[ Cu +2]

Q. $C u^{+ 2} + 2 e^{-} \to C u; lo g [C u^{+ 2}]$ vs $E_{RP}$ graph is of the type as shown in figure where $O A = 0.34$ volt. Then electrode potential of the half cell of $C u / C u^{+ 2} (0.1 M)$ will be:-

$(- 0.34 + \frac{0.0591}{2})$ volt

$(0.34 + 0.0591)$ volt

$0.34$ volt

None of these

$E_{RP} = E_{RP}^{o} - \frac{0.0591}{n} lo g_{10} \frac{1}{[ C u ^{+ 2} ]}$
$E_{RP} = E_{RP}^{o} + \frac{0.0591}{n} lo g_{10} [C u^{+ 2}]$
$y = c + m x$
$E_{RP}^{o} = 0.34 V$
$E_{OP}^{o} = - 0.34 V$
$E_{OP} = E_{OP}^{o} - \frac{0.0591}{2} lo g_{10} [C u^{+ 2}]$

Q. Cu+2+2e−→Cu;log[Cu+2] vs ERP​ 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(0.1M) will be:-

(−0.34+20.0591​) volt

(0.34+0.0591) volt

0.34 volt