
What will be the standard cell potential of the galvanic cell with the following reaction?
$2C{{r}_{(s)}}+3C{{d}^{2+}}_{(aq)}\to 2C{{r}^{3+}}_{(aq)}+3C{{d}_{(s)}}$
[Given: $E{{{}^\circ }_{C{{r}^{3+}}/Cr}}=-0.74V$ and $E{{{}^\circ }_{C{{d}^{2+}}/Cd}}=-0.40V$]
(A) $0.74V$
(B) $1.14V$
(C) $0.34V$
(D) $-0.34V$
Answer
542.7k+ views
Hint: In a redox reaction the oxidation and reduction of the constituent species takes place simultaneously, inside a galvanic cell.
The oxidation takes place in anode and the reduction takes place at cathode, in a galvanic cell.
Formula Used:$E{{{}^\circ }_{cell}}=E{{{}^\circ }_{Cathode}}-E{{{}^\circ }_{Anode}}$
Complete step-by-step solution:
In a galvanic cess we know that the standard electrode potential of a cell is equal to the subtraction of standard electrode potential of anode from the standard electrode potential of cathode. This equation can be represented by the following,
$E{{{}^\circ }_{cell}}=E{{{}^\circ }_{Cathode}}-E{{{}^\circ }_{Anode}}$
Here $E{{{}^\circ }_{cell}}$ signifies the standard electrode potential of the cell, the symbol $E{{{}^\circ }_{Cathode}}$ represents the standard electrode potential of cathode, and $E{{{}^\circ }_{Anode}}$ symbolises the standard electrode potential of anode.
Now if we consider the question, an equation is provided to us which is,
$2C{{r}_{(s)}}+3C{{d}^{2+}}_{(aq)}\to 2C{{r}^{3+}}_{(aq)}+3C{{d}_{(s)}}$
Here we can see that the chromium is being oxidised to a $+3$ state hence it acts as an anode, and the cadmium is reduced from a $+2$ to $0$ state, hence acting as a cathode.
The reduction potentials of both chromium and cadmium ion is provided to us in the question, now we will use these values in order to find out the standard cell potential of the galvanic cell.
Here the value of $E{{{}^\circ }_{Cathode}}$ would be $-0.40V$ and the values of $E{{{}^\circ }_{Anode}}$ would be $-0.74V$ as per the question. So, now we will put this values in the previous equation,
$E{{{}^\circ }_{cell}}=-0.40V-(-0.74V)$
Here we can see the negative signs would cancel out and the resultant value would be $0.34V$.
Now if we consider the options given in the questions, the value of our answer seems to match with option C.
So the correct option would be option (C).
Note: The standard cell potential of a galvanic cell is the difference between the standard reduction potential of cathode and the standard reduction potential of anode.
The species whose oxidation state increases during a redox reaction, are the ones which get oxidised, and the species whose oxidation state decreases, or becomes more negative, are the ones which get reduced.
The oxidation takes place in anode and the reduction takes place at cathode, in a galvanic cell.
Formula Used:$E{{{}^\circ }_{cell}}=E{{{}^\circ }_{Cathode}}-E{{{}^\circ }_{Anode}}$
Complete step-by-step solution:
In a galvanic cess we know that the standard electrode potential of a cell is equal to the subtraction of standard electrode potential of anode from the standard electrode potential of cathode. This equation can be represented by the following,
$E{{{}^\circ }_{cell}}=E{{{}^\circ }_{Cathode}}-E{{{}^\circ }_{Anode}}$
Here $E{{{}^\circ }_{cell}}$ signifies the standard electrode potential of the cell, the symbol $E{{{}^\circ }_{Cathode}}$ represents the standard electrode potential of cathode, and $E{{{}^\circ }_{Anode}}$ symbolises the standard electrode potential of anode.
Now if we consider the question, an equation is provided to us which is,
$2C{{r}_{(s)}}+3C{{d}^{2+}}_{(aq)}\to 2C{{r}^{3+}}_{(aq)}+3C{{d}_{(s)}}$
Here we can see that the chromium is being oxidised to a $+3$ state hence it acts as an anode, and the cadmium is reduced from a $+2$ to $0$ state, hence acting as a cathode.
The reduction potentials of both chromium and cadmium ion is provided to us in the question, now we will use these values in order to find out the standard cell potential of the galvanic cell.
Here the value of $E{{{}^\circ }_{Cathode}}$ would be $-0.40V$ and the values of $E{{{}^\circ }_{Anode}}$ would be $-0.74V$ as per the question. So, now we will put this values in the previous equation,
$E{{{}^\circ }_{cell}}=-0.40V-(-0.74V)$
Here we can see the negative signs would cancel out and the resultant value would be $0.34V$.
Now if we consider the options given in the questions, the value of our answer seems to match with option C.
So the correct option would be option (C).
Note: The standard cell potential of a galvanic cell is the difference between the standard reduction potential of cathode and the standard reduction potential of anode.
The species whose oxidation state increases during a redox reaction, are the ones which get oxidised, and the species whose oxidation state decreases, or becomes more negative, are the ones which get reduced.
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