The standard electrode potential of Zn2+/Zn is - 0.76 V and that of Cu2+/Cu is 0.34 V. The emf (V) and the free energy change (kJ mol-1 ), respectively for a daniell cell will be

Question

The standard electrode potential of Zn2+/Zn is - 0.76 V and that of Cu2+/Cu is 0.34 V. The emf (V) and the free energy change (kJ mol-1 ), respectively for a daniell cell will be (A) - 0.42 and 81 (B) 1.1 and - 213 (C) - 1.1 and 213 (D) 0.42 and -81

Tardigrade Admin · Accepted Answer

Given, E Zn 2+ / Zn °=-0.76 V E Cu 2+ / Cu ° =0.34 V E text cell ° =0.34-(-0.76) =1.1 V Δ G°=-n F E text cell ° =-2 × 96500 × 11 =-213 V

Q. The standard electrode potential of $Z n^{2 +} / Z n$ is $- 0.76 V$ and that of $C u^{2 +} / C u$ is $0.34 V$ . The emf $(V)$ and the free energy change $(k J m o l^{- 1})$ , respectively for a daniell cell will be

- 0.42 and 81

1.1 and - 213

- 1.1 and 213

0.42 and -81

Given, $E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V$
$E_{C u^{2 +} / C u}^{\circ} = 0.34 V$
$E_{cell}^{\circ} = 0.34 - (- 0.76)$
$= 1.1 V$
$Δ G^{\circ} = - n F E_{cell}^{\circ}$
$= - 2 \times 96500 \times 11$
$= - 213 V$

Q. The standard electrode potential of Zn2+/Zn is −0.76V and that of Cu2+/Cu is 0.34V. The emf (V) and the free energy change (kJmol−1), respectively for a daniell cell will be