Question

The e.m.f. of a cell, whose half-cells are given below, is: $M{{g}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mg(s);$ $E{}^\circ =-\,2.37\,\,V$ $C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu(s);$ $E{}^\circ =+\,0.34\,\,V$

• A. $+1.36\,\,V$
• B. $+\,2.71\,V$
• C. $-\,2.03\,\,V$
• D. $-\,2.71\,\,V$

Answer 1

Answer: (B)

$+\,2.71\,V$

Answer 2

In the cell Mg will act as anode while Cu will act as cathode as the reduction potential of Mg is less Hence, $E_{cell}^{\text{o}}=E_{cathode}^{\text{o}}-E_{anode}^{\text{o}}$ $=\left( 0.34 \right)-\left( -\,2.37 \right)$ $=0.34+2.37=2.71\,\,V$