Answer
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Hint: The answer for this question includes that when two electrodes are combined the one gets reduced at cathode and other gets oxidised at anode and thus the reduction potential is calculated using formula${{E}^{0}}_{cell}=({{E}^{0}}_{c}-{{E}^{0}}_{a})$
Complete step – by – step solution:
We have come across the several electrodes and their functions in the electrochemistry part that is in the physical chemistry part. Now, here in the above question the given cells are combined which means that there are two cells that are one anode and the other cathode.
We know that at cathode only reduction takes place always where in the reduction potential increases and at anode oxidation takes place where the reduction potential decreases.
Based on these facts, the cell potential can easily be calculated by,
${{E}^{0}}_{cell}=({{E}^{0}}_{c}-{{E}^{0}}_{a})$
From the data given we have, ${{E}^{0}}_{c}=+0.25V$that is for cathode and for anode${{E}^{0}}_{a}=-2.87V$
Substituting these values, ${{E}^{0}}_{cell}=\left[ 0.25-\left( -2.87 \right) \right]=0.25+2.87$
\[\Rightarrow {{E}^{0}}_{cell}=+3.12V\]
Now, since the reduction potential value of the reference electrode is zero and doesn’t play any role in this, the standard reduction potential of the individual as well as the combined cell remains independent from the reference electrode.
Therefore, the correct answer is option (B) 3.12V and independent on the reference electrode chosen.
Note: Here make sure that you are familiar with the notations given for some electrodes. Here, standard hydrogen electrode is also given as SHE which does not include any full stop symbols and if questions are based on this, do not be confused.
Complete step – by – step solution:
We have come across the several electrodes and their functions in the electrochemistry part that is in the physical chemistry part. Now, here in the above question the given cells are combined which means that there are two cells that are one anode and the other cathode.
We know that at cathode only reduction takes place always where in the reduction potential increases and at anode oxidation takes place where the reduction potential decreases.
Based on these facts, the cell potential can easily be calculated by,
${{E}^{0}}_{cell}=({{E}^{0}}_{c}-{{E}^{0}}_{a})$
From the data given we have, ${{E}^{0}}_{c}=+0.25V$that is for cathode and for anode${{E}^{0}}_{a}=-2.87V$
Substituting these values, ${{E}^{0}}_{cell}=\left[ 0.25-\left( -2.87 \right) \right]=0.25+2.87$
\[\Rightarrow {{E}^{0}}_{cell}=+3.12V\]
Now, since the reduction potential value of the reference electrode is zero and doesn’t play any role in this, the standard reduction potential of the individual as well as the combined cell remains independent from the reference electrode.
Therefore, the correct answer is option (B) 3.12V and independent on the reference electrode chosen.
Note: Here make sure that you are familiar with the notations given for some electrodes. Here, standard hydrogen electrode is also given as SHE which does not include any full stop symbols and if questions are based on this, do not be confused.
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