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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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