chapter 5 electrochemistry class 12 chemistry textbook solution
4. Answer the following :
xiv. Predict whether the following reactions would occur spontaneously under standard state conditions.
a. Ca (s) + Cd2⊕ (aq) →Ca2⊕(aq) + Cd(s)
b. 2 Br (s) + Sn2⊕ (aq) →Br2(l) + Sn(s)
c. 2Ag(s) + Ni2⊕ (aq)→ 2 Ag⊕ (aq) + Ni (s)
(use information of Table 5.1)
Answer:-
Ca(s) + Cd2+(aq) → Ca2+(aq) + Cd(s)
At the anode: Ca(s) → Ca2+(aq) + 2e-
At the cathode: Ca2+(aq) + 2e- → Cd(s)
From the electrochemical series we have,
ECacirc = -2.866 V and ECdcirc = -0.403 V
For the cell having Ca as the anode and Cd as the cathode.
Ecellcirc = ECdcirc - ECacirc
= -0.403 V - (-2.866 V)
Ecellcirc = 2.463 V
Since the emf of the cell is positive, the given cell reaction is spontaneous.
2Br-(aq) + Sn2+(aq) → Br2(l) + Sn(s)
At the anode: 2Br-(aq) → Br2(l) + 2e-
At the cathode: Sn2+(aq) + 2e- → Sn(s)
From the electrochemical series, we have,
EBr2circ = 1.080 V and ESncirc = -0.136 V
For the cell having Br2(l) as the anode and Sn as the cathode.
Ecellcirc = ESncirc - EBr2circ
= -0.136 V - 1.080 V
Ecellcirc = -1.216 V
Since the emf of the cell is negative, the given cell reaction is nonspontaneous.
\[ \ce{2Ag_{(s)} + Ni^{2+}_{(aq)} -> 2Ag^{+}_{(aq)} + Ni_{(s)}} \]
At the anode:
\[ \ce{2Ag_{(s)} -> 2Ag^+_{(aq)} + 2e^-} \]
At the cathode:
\[ \ce{Ni^{2+}_{(aq)} + 2e^- -> Ni_{(s)}} \]
From the electrochemical series, we have,
\(E^\circ_{\text{Ag}}\) = 0.799 V and \(E^\circ_{\text{Ni}}\) = -0.257 V
For the cell having Ag as the anode and Ni as the cathode:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ni}} - E^\circ_{\text{Ag}} \]
= -0.257 V - 0.799 V
\(E^\circ_{\text{cell}}\) = -1.056 V
Since the emf of the cell is negative, the given cell reaction is nonspontaneous.
chapter 5 elctrochemistry textbook solution page 118