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What is the relationship between concentration and voltage in a galvanic cell? How do you calculate the theoretical?
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Larry Nelson
What is the relationship between concentration and voltage in a galvanic cell? How do you calculate the theoretical?
The potential of the Copper electrode is given by the Nernst equation : $$E(Cu^{2+}/Cu) = + 0.34 V + 0.0295 log[Cu^{2+}]$$ If the concentration $[Cu^{2+}]$ is $0.25 M$ and then $0.50 M$, the potential of these half-cells become respectively $0.32$ V and $0.33 V$.
If you couple this half-cell with a Zinc electrode with $[Zn^{2+}]$ = $0.1 M$, the same sort of calculation gives you a potential $E(Zn^{2+}/Zn)$ =$ -0.76 + 0.0295(-1)$ = $-0.79 V$. If you combine both half-cells, you obtain $1.11 V$ and $1.12 V$ respectively.
This being said, I would like to advise you not to use Zinc nitrate in the zinc half-cell. Please replace Zinc nitrate by Zinc sulfate. The reason is that the ion nitrate attacks the Zinc metal, producing nitrite ions. So the potential of the half-cell made with Zinc metal in a Zinc nitrate solution is not stable.
The potential of the Copper electrode is given by the Nernst equation : $$E(Cu^{2+}/Cu) = + 0.34 V + 0.0295 log[Cu^{2+}]$$ If the concentration $[Cu^{2+}]$ is $0.25 M$ and then $0.50 M$, the potential of these half-cells become respectively $0.32$ V and $0.33 V$.
If you couple this half-cell with a Zinc electrode with $[Zn^{2+}]$ = $0.1 M$, the same sort of calculation gives you a potential $E(Zn^{2+}/Zn)$ =$ -0.76 + 0.0295(-1)$ = $-0.79 V$. If you combine both half-cells, you obtain $1.11 V$ and $1.12 V$ respectively.
This being said, I would like to advise you not to use Zinc nitrate in the zinc half-cell. Please replace Zinc nitrate by Zinc sulfate. The reason is that the ion nitrate attacks the Zinc metal, producing nitrite ions. So the potential of the half-cell made with Zinc metal in a Zinc nitrate solution is not stable.
As a side comment, the concentration of the cupric may be impacted by the equilibrium reaction: Cu + Cu(II) = 2 Cu(l) . This reaction is claimed to be accelerated at lower pH and the nature of the copper salt to undergo hydrolysis may be impactful.More
The potential of the Copper electrode is given by the Nernst equation : $$E(Cu^{2+}/Cu) = + 0.34 V + 0.0295 log[Cu^{2+}]$$ If the concentration $[Cu^{2+}]$ is $0.25 M$ and then $0.50 M$, the potential of these half-cells become respectively $0.32$ V and $0.33 V$.
If you couple this half-cell with a Zinc electrode with $[Zn^{2+}]$ = $0.1 M$, the same sort of calculation gives you a potential $E(Zn^{2+}/Zn)$ =$ -0.76 + 0.0295(-1)$ = $-0.79 V$. If you combine both half-cells, you obtain $1.11 V$ and $1.12 V$ respectively.
This being said, I would like to advise you not to use Zinc nitrate in the zinc half-cell. Please replace Zinc nitrate by Zinc sulfate. The reason is that the ion nitrate attacks the Zinc metal, producing nitrite ions. So the potential of the half-cell made with Zinc metal in a Zinc nitrate solution is not stable.
The potential of the Copper electrode is given by the Nernst equation : $$E(Cu^{2+}/Cu) = + 0.34 V + 0.0295 log[Cu^{2+}]$$ If the concentration $[Cu^{2+}]$ is $0.25 M$ and then $0.50 M$, the potential of these half-cells become respectively $0.32$ V and $0.33 V$.
If you couple this half-cell with a Zinc electrode with $[Zn^{2+}]$ = $0.1 M$, the same sort of calculation gives you a potential $E(Zn^{2+}/Zn)$ =$ -0.76 + 0.0295(-1)$ = $-0.79 V$. If you combine both half-cells, you obtain $1.11 V$ and $1.12 V$ respectively.
This being said, I would like to advise you not to use Zinc nitrate in the zinc half-cell. Please replace Zinc nitrate by Zinc sulfate. The reason is that the ion nitrate attacks the Zinc metal, producing nitrite ions. So the potential of the half-cell made with Zinc metal in a Zinc nitrate solution is not stable.
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