Calculate the emf of the cell in which of the following reaction takes place Ni (s)+2 Ag +(0.002 M ) longrightarrow Ni 2+(0.160 M ) + 2Ag(s) (Give that E text cell °=1.05 V )

Question

Calculate the emf of the cell in which of the following reaction takes place Ni (s)+2 Ag +(0.002 M ) longrightarrow Ni 2+(0.160 M ) + 2Ag(s) (Give that E text cell °=1.05 V ) (A) 0.73 V (B) 0.91 V (C) 0.62 V (D) 0.34 V

Tardigrade Admin · Accepted Answer

From the given cell reaction and Nernst equation,

E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} lo g \frac{[ N i ^{2 +} ]}{[ A g ^{+} ] ^{2}}

= 1.05 V - \frac{0.0591}{2} lo g \frac{[ 0.160 ]}{[ 0.002 ] ^{2}}

= 1.05 - \frac{0.0591}{2} lo g (4 \times 1 0^{4})

= 1.05 - \frac{0.0591}{2} (4.6021)

= 1.05 - 0.14 = + 0.91 V

E_{cell} = + 0.91 V

Q. Calculate the emf of the cell in which of the following reaction takes place Ni(s)+2Ag+(0.002M)⟶Ni2+(0.160M)+2Ag(s) (Give that Ecell ∘​=1.05V)

0.73 V

0.91 V

0.62 V

0.34 V

From the given cell reaction and Nernst equation, Ecell ​=Ecell ∘​−n0.0591​log[Ag+]2[Ni2+]​ =1.05V−20.0591​log[0.002]2[0.160]​ =1.05−20.0591​log(4×104) =1.05−20.0591​(4.6021) =1.05−0.14=+0.91V Ecell ​=+0.91V

Q. Calculate the emf of the cell in which of the following reaction takes place
$N i (s) + 2 A g^{+} (0.002 M) ⟶ N i^{2 +} (0.160 M) + 2 A g (s)$
(Give that $E_{cell}^{\circ} = 1.05 V)$