Which expression gives the correct value for the standard potential for a gold-rhodium voltaic cell?
Half-reaction
E∘(V)
Rh+3(aq)+3e−→Rh(s)
0.76V
Au+(aq)+e−→Au(s)
1.69V
1.69V−0.76V
1.69V+0.76V
3(1.69)V+0.76V
3(1.69)V−0.76V
A
1.69V+0.76V
B
3(1.69)V+0.76V
C
3(1.69)V−0.76V
D
1.69V−0.76V
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Solution
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We know that, metal having higher reduction potential undergoes reduction;
so, Au will undergo reduction and Rh will undergo oxidation.
E0cell = 1.69 V - 0.76 V
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Similar Questions
Q1
Which expression gives the correct value for the standard potential for a gold-rhodium voltaic cell?
Half-reaction
E∘(V)
Rh+3(aq)+3e−→Rh(s)
0.76V
Au+(aq)+e−→Au(s)
1.69V
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Q2
Which is the correct value for the standard potential for a gold-rhodium voltaic cell?
The standard half cell potential at 298K are: Rh3+(aq)+3e−→Rh(s);Eored=0.76V Au+(aq)+e−→Au(s);Eored=1.69V
View Solution
Q3
For the following reactions: Zn→Zn2++2e−;E=+0.76V Au→Au3++3e−;E=−1.42V If gold foil is placed in a solution containing Zn2+, the reaction potential would be:
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Q4
For a reaction Zn+2+2e−⟶Zn,Eo=−0.76V which of the following statement is true? (E=Electrode potential)
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Q5
For Zn2+/Zn,E∘=−0.76V, for Ag+/Ag,E∘=0.799V. The correct statement is: