Standard reduction potentials for the half-reactions associated with the electrochemical cell shown above are given in the table below.
a. Which of the following is the net ionic equation for the overall reaction that occurs as the cell operates?
SOLUTION:
In the cell, there are two half reactions. Oxidation(loss of electrons) is on the left and reduction (gain of electrons) is on the right. Half-reactions given in the table are both for reduction. We need to flip the second half-reaction to get oxidation.
We get Zn(s) –> Zn2+(aq)+ 2e-
We also have Ag+(aq)+ e- –> Ag(s)
To balance the two half reaction so that the electrons can cancel out, we need to multiply the bottom one by 2. We get:
2Ag+(aq)+ 2e- –> 2Ag(s)
Zn(s) –> Zn2+(aq)+ 2e-
Adding the two half reactions together, electrons will cancel and we get the final equation:
2Ag+(aq)+ Zn(s) –> 2Ag(s)+Zn2+(aq), Choice D
b. What is the value of E° for the cell?
(A) 0.04 V (B) 0.84 V (C) 1.56 V (D) 2.36 V
SOLUTION:
E°cell = E° oxidation + E° reduction
In the chart we are given reduction potentials for both half-reactions. Since the bottom half reaction gets flipped, we need to change the sign of its potential.
Zn(s) –> Zn2+(aq)+ 2e- E° reduction = 0.76V
Ag+(aq)+ e- –> Ag(s) E° reduction =0.80V
E°cell = 0.76V+0.80V = 1.56V, choice C
LINKS
General Chemistry Course
Online General Chemistry Tutor
Find me on Youtube
Find me on TikTok
Learn more about Transformation Tutoring