Answer:
Explanation:
Step 1: Write both half reactions
Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-
Ni2+ / Ni : reduction Ni2+ +2e- → Ni
Step 2: Balance reactions and look up the standard potential for the half-reactions
2(Cr → Cr3+ + 3e-) E° ox = 0.74 V
3(Ni2+ +2e- → Ni) E° red = -0.25 V
2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni
E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49
E
=
E
°
−
0.0257
V
/n *
ln
Q =
E
°
−
0.0257
V
/n *l
n
[
C
r
3
+
]/
[
N
i
2
+
]
With E° = 0.49 V
b)
Step 1: Write both half reactions
MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O
Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-
Step 2: Balance reactions and look up the standard potential for the half-reactions
MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V
5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V
2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+
E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V
E
=
E
°
−
0.0257
V
/n *
ln
Q =
E
°
−
0.0257
V
/n *l
n
[
C
f2
+
]/
[
Mno4-
]
With E° =3.63 V
