Analyze and solve this partially completed galvanic cell puzzle. There are 4 ele
ID: 1038026 • Letter: A
Question
Analyze and solve this partially completed galvanic cell puzzle. There are 4 electrodes each identified by a letter of the alphabet, A through D. The values in the partially completed grid are measured cell potentials for a cell consisting of electrode #1 and electrode #2. You may assume that each galvanic cell was properly constructed with the appropriate metals and solutions and that all the measured values in the grid are accurate.
electrode #1 ?
C
B
D
A
electrode #2 ?
Ecell(volts)
Ecell(volts)
Ecell(volts)
Ecell(volts)
C
0
0.91
-0.62
B
-0.91
0
-1.53
-0.24
D
0.62
1.53
0
1.29
1.77 volts
1.53 volts
0.67 volts
0.38 volts
0.24 volts
1.15 volts
0 volts
2.20 volts
2.82 volts
1.91 volts
0.86 volts
1.05 volts
1.29 volts
electrode #1 ?
C
B
D
A
electrode #2 ?
Ecell(volts)
Ecell(volts)
Ecell(volts)
Ecell(volts)
C
0
0.91
-0.62
?B
-0.91
0
-1.53
-0.24
D
0.62
1.53
0
1.29
Explanation / Answer
Considering E(A), E(B), E(c) and E(D) to be the potentials corresponding to the four electrodes A, B, C and D respectively.
Let E(C)-E(B)=x (the standard cell potential that we need to calculate).................(1)
E(B)-E(A)=-0.24, i.e., E(A) = E(B) + 0.24...................(2)
Putting equation (2) in (1) we get,
E(C)-E(B)-0.24=x..................(3)
From the values in the grid provided we know, E(C)-E(B)=0.91.........................(4)
Putting equation (4) in (3) we get,
0.91-0.24=x or, x= 0.67
i.e., E(C)-E(A) =0.67
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