Question

The standard reduction potential $E^{\circ }$, for the half reaction are:-
$Zn\to Z{n}^{2+}+2e^{-};E^{\circ }=0.76V$
$Cu\to C{u}^{2+}+2e^{-};E^{\circ }=0.34V$
The emf for the cell reaction, $Zn\left(s\right)+C{u}^{2+}\to Z{n}^{2+}+Cu\left(s\right)$ is_____________.

• A. $0.42V$
• B. $-0.42V$
• C. $-1.1V$
• D. $+1.1V$

Answer 1

The correct option is D $+1.1V$

$Z{n}_{\left(aq\right)}^{2+}+2e^{-}\to Z{n}_{\left(s\right)}$ $E^{\circ }=0.76V$.

$C{u}_{\left(aq\right)}^{2+}+2e^{-}\to C{u}_{\left(s\right)}$ $E^{\circ }=0.34V$.

Half cell reaction of $Z{n}^{2+}/Zn$ electrode is reversed and the sign of $E^{\circ }$ is changed.

$\therefore Z{n}_{\left(s\right)}\to Z{n}_{\left(aq\right)}^{2+}+2e^{-}$ $E^{\circ }=-0.76V$.

$\therefore$ For emf of cell reaction:-

$Z{n}_{\left(s\right)}\to Z{n}^{2+}+2e^{-}$-------oxidation[anode].

$C{u}_{\left(aq\right)}^{2+}+2e^{-}\to C{u}_{\left(s\right)}$--------------reduction[cathode].

$\therefore E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$

$=0.34-(-0.76)=+1.10V$