The e.m.f. of a cell, whose half-cells are given below, is: Mg2++2e- xrightarrowMg(s); E° =- 2.37 V Cu2++2e- xrightarrowCu(s); E° =+ 0.34 V

Question

The e.m.f. of a cell, whose half-cells are given below, is: Mg2++2e- xrightarrowMg(s); E° =- 2.37 V Cu2++2e- xrightarrowCu(s); E° =+ 0.34 V  (A)  +1.36 V  (B)  + 2.71 V  (C)  - 2.03 V  (D)  - 2.71 V

Tardigrade Admin · Accepted Answer

In the cell Mg will act as anode while Cu will act as cathode as the reduction potential of Mg is less Hence, Ecell texto=Ecathode texto-Eanode texto =( 0.34 )-( - 2.37 ) =0.34+2.37=2.71 V

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

$+ 1.36 V$

$+ 2.71 V$

$- 2.03 V$

$- 2.71 V$

In the cell Mg will act as anode while Cu will act as cathode as the reduction potential of Mg is less Hence, $E_{ce ll}^{o} = E_{c a t h o d e}^{o} - E_{an o d e}^{o}$ $= (0.34) - (- 2.37)$ $= 0.34 + 2.37 = 2.71 V$

Q. The e.m.f. of a cell, whose half-cells are given below, is: Mg2++2e−​Mg(s); E∘=−2.37V Cu2++2e−​Cu(s); E∘=+0.34V

+1.36V

+2.71V

−2.03V

−2.71V