
For a cell reaction involving two electron change, the standard EMF of the cell is 0.295 V at 25$^\circ $C C. The equilibrium constant of the reaction at 25$^\circ $C C will be:
A. \[29.2 \times {10^{ - 2}}\]
B. \[10\]
C. \[1 \times {10^{10}}\]
D. \[2.95 \times {10^{ - 10}}\]
Answer
521.1k+ views
Hint: Initially you should be aware of the cell reaction and the formulas related to cell reaction and galvanic cell. You should also check the values and units of each point.
Complete step by step answer:
We have,
\[
E.M.F = 0.295V \\
T = {25^ \circ }C \\
\]
The formula to find \[{E_{cell}} ^\circ \] is as follows:
\[{E_{cell}} ^\circ = \dfrac{{2.303RT\log {K_c}}}{{nF}}\]
By placing the values and calculating, we get:
\[{E_{cell}} ^\circ = \dfrac{{0.0591 \times log{K_c}}}{n}\]
Since two electrons change is involved in the cell reaction therefore n=2
\[
0.259 = \dfrac{{0.0591 \times log{K_c}}}{2} \\
\log {K_c} = 10 \\
{K_c} = 1 \times {10^{10}} \\
\]
It is the required equilibrium constant of the reaction at 25$^\circ $CC .
So, the correct answer is Option C.
Note: The cell reaction is that the overall reaction which takes place within the cell, written on the idea that the proper hand electrode is that the cathode, i.e., assuming that the spontaneous reaction is that the one during which reduction is happening within the right-hand compartment.
To write the cell reaction like a cell diagram, the right-hand half reaction is written as a discount, and therefore the left-hand half-reaction, written as an oxidation, is added thereto. (This is strictly an equivalent as subtracting the left-hand equation written as a discount, which is that the formally correct procedure.)
Thus, the cell reaction for the cell described above is as follows: \[2F{e^{3 + }}_{(aq)}\; + {\text{ }}Cu{\;_{(s)}}\;{\text{ }} \to \;{\text{ }}2F{e^{2 + }}_{(aq)}\; + {\text{ }}C{u^{2 + }}_{(aq)}\]
In chemistry, conductor potential is the electrical phenomenon of a voltaic cell engineered from a regular reference conductor and another conductor to be characterised. By convention, the reference conductor is that the normal atomic number 1 conductor (SHE). it's outlined to own a possible of zero volts.
Complete step by step answer:
We have,
\[
E.M.F = 0.295V \\
T = {25^ \circ }C \\
\]
The formula to find \[{E_{cell}} ^\circ \] is as follows:
\[{E_{cell}} ^\circ = \dfrac{{2.303RT\log {K_c}}}{{nF}}\]
By placing the values and calculating, we get:
\[{E_{cell}} ^\circ = \dfrac{{0.0591 \times log{K_c}}}{n}\]
Since two electrons change is involved in the cell reaction therefore n=2
\[
0.259 = \dfrac{{0.0591 \times log{K_c}}}{2} \\
\log {K_c} = 10 \\
{K_c} = 1 \times {10^{10}} \\
\]
It is the required equilibrium constant of the reaction at 25$^\circ $CC .
So, the correct answer is Option C.
Note: The cell reaction is that the overall reaction which takes place within the cell, written on the idea that the proper hand electrode is that the cathode, i.e., assuming that the spontaneous reaction is that the one during which reduction is happening within the right-hand compartment.
To write the cell reaction like a cell diagram, the right-hand half reaction is written as a discount, and therefore the left-hand half-reaction, written as an oxidation, is added thereto. (This is strictly an equivalent as subtracting the left-hand equation written as a discount, which is that the formally correct procedure.)
Thus, the cell reaction for the cell described above is as follows: \[2F{e^{3 + }}_{(aq)}\; + {\text{ }}Cu{\;_{(s)}}\;{\text{ }} \to \;{\text{ }}2F{e^{2 + }}_{(aq)}\; + {\text{ }}C{u^{2 + }}_{(aq)}\]
In chemistry, conductor potential is the electrical phenomenon of a voltaic cell engineered from a regular reference conductor and another conductor to be characterised. By convention, the reference conductor is that the normal atomic number 1 conductor (SHE). it's outlined to own a possible of zero volts.
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