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(a)
Interpretation:
The cell potential is to be calculated.
Concept introduction:
Reduction will occur at that electrode having high reduction potential.
Answer to Problem 41A
The given cell reaction is2Ag+2+Pb⇌Pb+2+2AgThe cell representation of given cell reaction isAg+1|Ag‖Pb+2|PbThe cell potential is to be calculatedE°cell=E°cathode-E°anode =0.7996-(-0.1262) =0.9258 V
The cell potential is 0.9258 V.
Explanation of Solution
Reduction will occur at that electrode having high reduction potential.
The given cell reaction is2Ag+2+Pb⇌Pb+2+2AgThe cell representation of given cell reaction isAg+1|Ag‖Pb+2|PbThe cell potential is to be calculatedE°cell=E°cathode-E°anode =0.7996-(-0.1262) =0.9258 V
(b)
Interpretation:
The cell potential is to be calculated.
Concept introduction:
Reduction will occur at that electrode having high reduction potential
(b)
Answer to Problem 41A
The given cell reaction isMn(s)+Ni+2(aq)⇌Mn+2+NiThe cell representation of given cell reaction isMn|Mn+2‖Ni+2|NiThe cell potential is to be calculatedE°cell=E°cathode-E°anode =(-0.257)-(-1.185) =0.928 V
The cell potential is 0.928 V
Explanation of Solution
Reduction will occur at that electrode having high reduction potential.
The given cell reaction isMn(s)+Ni+2(aq)⇌Mn+2+NiThe cell representation of given cell reaction isMn|Mn+2‖Ni+2|NiThe cell potential is to be calculatedE°cell=E°cathode-E°anode =(-0.257)-(-1.185) =0.928 V
The cell potential is 0.928 V
(c)
Interpretation:
The cell potential is to be calculated.
Concept introduction:
Reduction will occur at that electrode having high reduction potential.
(c)
Answer to Problem 41A
The given cell reaction isI2(aq)+ Sn(aq)⇌2I-+ Sn+2The cell representation of given cell reaction isSn|Sn+2‖I-1|I2The cell potential is to be calculatedE°cell=E°cathode-E°anode =(0.5355)-(-0.1375) =0.6730 V
The cell potential is 0.673 V
Explanation of Solution
The standard reduction potential of Sn|Sn+2 is greater than the reduced potential of iodine ion. So, reduction is occurs at Sn|Sn+2 electrode and oxidation is occurs at I-1|I2 electrode.
The given cell reaction isI2(aq)+ Sn(aq)⇌2I-+ Sn+2The cell representation of given cell reaction isSn|Sn+2‖I-1|I2The cell potential is to be calculatedE°cell=E°cathode-E°anode =(0.5355)-(-0.1375) =0.6730 V
The cell potential is 0.673 V.
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