then which one is not correct
- A$E = 0$
- B$\frac{{RT}}{{nF}}\,\ln {K_{eq}} = {E^o}$
- ✓$E = {E^o}$
- D${K_{eq}} = {e^{\frac{{n{E^o}F}}{{RT}}}}$
then which one is not correct
$E = 0;{E^o} \ne 0$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
(Assume $100\%$ ionization)
$A.$ $0.500\,M\,C _2 H _5 OH ( aq )$ and $0.25\, M\, KBr ( aq )$
$B.$ $0.100\,M\,K _4\left[ Fe ( CN )_6\right]$ (aq) and $0.100\, M$ $FeSO _4\left( NH _4\right)_2 SO _4$ (aq)
$C.$ $0.05 \,M\, K _4\left[ Fe ( CN )_6\right]( aq )$ and $0.25\, M\, NaCl$ (aq)
$D.$ $0.15\, M\, NaCl ( aq )$ and $0.1\, M BaCl _2$ (aq)
$E.$ $0.02\, M\, KCl\, MgCl _{2 .} 6 H _2 O ( aq )$ and $0.05\, M$ $KCl ( aq )$

$Mg\left( s \right) + \mathop {2A{g^ + }}\limits_{(0.0001\,M)} \to \mathop {M{g^{ + 2}}}\limits_{\left( {0.100\,M} \right)} + 2Ag\left( s \right)$
If $E_{cell}^o = 3.17\,V$