Consider the following four electrodes:
P = Cu²⁺ (0.0001 M)/Cu₍ₛ₎
Q = Cu²⁺ (0.1 M)/Cu₍ₛ₎
R = Cu²⁺ (0.01 M)/Cu₍ₛ₎
S = Cu²⁺ (0.001 M)/Cu₍ₛ₎
If the standard reduction potential of Cu²⁺/Cu is +0.34 V,
the reduction potentials in volts of the above electrodes
follow the order.
(a) P > S > R > Q (b) S > R > Q > P
(c) R > S > Q > P (d) Q > R > S > P
Answers
Answer:
Consider the following four electrodes:
P = Cu²⁺ (0.0001 M)/Cu₍ₛ₎
Q = Cu²⁺ (0.1 M)/Cu₍ₛ₎
R = Cu²⁺ (0.01 M)/Cu₍ₛ₎
S = Cu²⁺ (0.001 M)/Cu₍ₛ₎
If the standard reduction potential of Cu²⁺/Cu is +0.34 V,
the reduction potentials in volts of the above electrodes
follow the order.
(a) P > S > R > Q
(b) S > R > Q > P
(c) R > S > Q > P
(d) Q > R > S > P
The reduction potentials in volts of the electrodes follow the order as :
• The standard reduction potential of Cu²⁺/Cu is +0.34 V
• Given four electrodes :
P = Cu²⁺ (0.0001 M)/Cu₍ₛ₎
Q = Cu²⁺ (0.1 M)/Cu₍ₛ₎
R = Cu²⁺ (0.01 M)/Cu₍ₛ₎
S = Cu²⁺ (0.001 M)/Cu₍ₛ₎
We know that,
• When concentration of Cu²⁺ increased, equilibrium will shift to left side and making the reduction potential more positive.
• Reduction potential becomes more positive when concentration is decreased.
• Concentration order,
Q > R > S > P
• Hence correct order of decrease in reduction potential is
Q > R > S > P