The standard reduction potential Eo for the half reactions are as: Z n arrow Z n2++2 e-, Eo=0.76 V C u arrow C u2++2 e-, Eo=0.34 V The emf for the cell reaction: Z n+C u2+ arrow Z n2+C u

Question

The standard reduction potential Eo for the half reactions are as: Z n arrow Z n2++2 e-, Eo=0.76 V C u arrow C u2++2 e-, Eo=0.34 V The emf for the cell reaction: Z n+C u2+ arrow Z n2+C u (A)  0.42 V  (B)  -0.42 V  (C)  -1.1 V  (D)  1.1 V

Tardigrade Admin · Accepted Answer

(i)Decide cathode and anode (ii) Ec e l lo=Eo C-Eo A of electrode potential are substituted in terms of reduction potential. Given, EZ n / Z n2+o=0.76 V EC u / C u2+o=0.34 V ∴ Zn is anode ( ∵ it has higher oxidation potential) ∴ EZ n2+ / Z no=-0.76 V and EC u2+ / C uo=-0.34 V Ec e l lo=Eo C-Eo A =-0.34 V-(-0.76 V) =0.34 V+0.76 V =0.42 V

Q. The standard reduction potential $E^{o}$ for the half reactions are as:
$Z n \to Z n^{2 +} + 2 e^{-}, E^{o} = 0.76 V$
$C u \to C u^{2 +} + 2 e^{-}, E^{o} = 0.34 V$
The emf for the cell reaction: $Z n + C u^{2 +} \to Z n^{2} + C u$

$0.42 V$

$- 0.42 V$

$- 1.1 V$

$1.1 V$

Q. The standard reduction potential Eo for the half reactions are as: Zn→Zn2++2e−,Eo=0.76V Cu→Cu2++2e−,Eo=0.34V The emf for the cell reaction: Zn+Cu2+→Zn2+Cu

0.42V

−0.42V

−1.1V

1.1V

Q. The standard reduction potential $E^{o}$ for the half reactions are as:
$Z n \to Z n^{2 +} + 2 e^{-}, E^{o} = 0.76 V$
$C u \to C u^{2 +} + 2 e^{-}, E^{o} = 0.34 V$
The emf for the cell reaction: $Z n + C u^{2 +} \to Z n^{2} + C u$

$0.42 V$

$- 0.42 V$

$- 1.1 V$

$1.1 V$