The given electrodes and their standard reduction potentials are:
- A: Ag+(0.0001M)/Ag(s)
- B: Ag+(0.1M)/Ag(s)
- C: Ag+(0.01M)/Ag(s)
- D: Ag+(0.001M)/Ag(s), EAg+/Ag∘=+0.80 V
Explanation:
The reduction potential for the given electrodes can be calculated using the Nernst equation: E=E∘−n0.0591log([Oxidant][Reductant]) Since the number of electrons involved in the silver (Ag) reduction reaction is 1 (i.e., n=1), the equation simplifies to: E=E∘−0.0591log[Ag+] The standard reduction potential EAg+/Ag∘ is given as +0.80 V for standard conditions.
Step 1: Calculating Reduction Potentials:
Using the Nernst equation: - For electrode A: EA=0.80−0.0591log(0.0001)=0.80−0.0591×(−4)=0.80+0.2364=1.0364 V - For electrode B: EB=0.80−0.0591log(0.1)=0.80−0.0591×(−1)=0.80+0.0591=0.8591 V - For electrode C: EC=0.80−0.0591log(0.01)=0.80−0.0591×(−2)=0.80+0.1182=0.9182 V - For electrode D: ED=0.80−0.0591log(0.001)=0.80−0.0591×(−3)=0.80+0.1773=0.9773 V
Conclusion:
The reduction potentials for the electrodes in descending order are: B>C>D>A Therefore, the correct order is Option (A): B > C > D > A.
Correct Answer: Option (A)