The standard reduction potential data at 25ºC is given below.
Eº (Fe 3+ .Fe 2+ ) = + 0.77 V; Eº (Fe 2+ .Fe) = – 0.44 V;
Eº (Cu 2+ .Cu) = + 0.34 V; Eº (Cu + .Cu) = + 0.52 V;
Eº (O 2 (g) + 4H + + 4e – → 2H 2 O) = + 1.23 V ; Eº (O 2 (g) + 2H 2 O + 4e – → 4OH) = + 0.40 V
Eº (Cr 3+ .Cr) = – 0.74 V; Eº (Cr 2+ .Cr) = – 0.91 V
Match Eº of the rebox pair in List I with the values given in List II and select the correct answer using the code given below the lists :
List I | List II | ||
P. | Eº (Fe3+, Fe) | 1. | – 0.36 V |
Q. | Eº (4H2O 4H+ + 4OH–) | 2. | –0.4 V |
R. | Eº (Cu2+ + Cu → 2Cu+) | 3. | –0.04 V |
S. | Eº(Cr3+, Cr+2) | 4. | –0.83 V |
Codes
P Q R S
Text Solution
Verified by ExpertsD
(P) Eº Fe3+, Fe 
⇒ 1 × 0.77 + 2 × (– 0.44) = 3 × x
⇒ x = – 
V ~ – 0.04 V.
(Q) 4H 2 O 
4H + + 4OH –
2H 2 O ⎯→ O 2 + 4H + + 4e – – 1.23 V
+ O 2 + 2H 2 O + 4e – ⎯→ 4OH – + 0.4 V
4H 2 O 
4H + + 4OH – – 0.83 V
(R) Eº (Cu2+ + Cu → 2Cu+) 
x × 1 + 0.52 × 1 = 0.34 × 2
x = 0.16 V.
⇒ Cu 2+ + e – ⎯→ Cu + 0.16 V
+ Cu ⎯→ Cu + + e – – 0.52 V
Cu 2+ + Cu 
2Cu + – 0.36 V
However, in the given option, – 0.18 V is printed.
(s) Eº (Cr3+, Cr2+) 
x × 1 + 2 × (– 0.91) = 3 × (– 0.74)
x – 1.82 = – 2.22
⇒ x = – 0.4 V
Hence, most appropriate is .
(P) – ; (Q) – ; (R) – ; (S) – 2.
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