An electrochemical cell is written in such a way that the anode is written first

Question

An electrochemical cell is written in such a way that the anode is written first (on the left) followed by the cathode. Therefore, Ni/Ni2+ represents the anode. Oxidation occurs at the anode; therefore, Ni is oxidized to Ni2+ at the anode. The anode half reaction is

Oxidation: Ni (s) -------> Ni2+ (aq) + 2 e- ; E0A = +0.25 V

The standard reduction potential of Ni2+/Ni is listed in tables as -0.25 V; oxidation is the reverse process of reduction and hence the oxidation potential is the inverse of the standard reduction potential.

The value +0.257 V is the closest value and hence is the answer.

The right side represents the cathode reaction. Reduction takes place at the cathode. The cathode reaction is

Reduction: M4+ (aq, 0.025 M) + e- --------> M3+ (aq, 0.010 M)

The balanced redox reaction is

Ni (s) + 2 M4+ (aq, 0.025 M) --------> Ni2+ (aq, 0.25 M) + 2 M3+ (aq, 0.010 M)

Q = [Ni2+][M3+]2/[M4+]2 = (0.025 M)*(0.010 M)2/(0.025 M)2

= 4.0*10-3 M

Since Q is a reaction quotient, ignore the unit and report the value as Q = 0.004 (ans).

Explanation / Answer

Use the standard reduction potential values your chemistry data sheet to answer this question. The following electrochemical cell has a potential of E1.883 va from 298 K: 4+ 3+ Pt(s) Where M3 and M are ions of an unknown metal. First, determine the standard anode potential (EA) O-0.257.v O +0,257 v O 0.00 v O None of the above What is the value of Q at the cathode? HINT: Make sure you have balanced the overall reaction.. O 0.16 O 6.25 O 16 O 0.004

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