For my electrochemistry lab I'm using $\pu{10 mL}$ of $\ce{ZnSO4}$ and $\pu{10 mL}$ of $\ce{CuSO4}$ solutions connected with a salt bridge. I connect $\ce{Zn}$ metal and $\ce{Cu}$ metal to a voltmeter and dip those metals in the correct beakers. The voltage I get is $\pu{1.044 V}.$

However when I add $\pu{10 mL}$ of water to both beakers, I still receive the same result. Does diluting two solutions affect the voltage?

  • $\begingroup$ Do you have reason not to trust your experimental results? $\endgroup$ – ericksonla Jun 1 '18 at 17:44
  • 2
    $\begingroup$ What result were you expecting? What does the Nernst equation say? $\endgroup$ – Ivan Neretin Jun 1 '18 at 17:58

Cell potential definitely depends on the concentration of the two solutions. This dependence of the cell potential on the concentration can be described by the Nernst equation.

For the reaction you are suggesting the half cells are:

$$ \begin{align} \ce{Cu^2+ + 2 e- &-> Cu} &\quad E_1^\circ &= \pu{0.34 V}\tag{R1}\\ \ce{Zn^2+ + 2 e- &-> Zn} &\quad E_2^\circ &= \pu{-0.76 V}\tag{R2}\\ \end{align} $$

Since zinc is more reactive than copper, or since $\ce{Zn^2+}$ reduction is less thermodynamically favored than the reduction of $\ce{Cu^2+}$, then the half-equation of the reduction of $\ce{Zn^2+}$ must be reversed and so does the sign of its potential.

Combining the half equations gives:

$$\ce{Zn + Cu^2+ -> Zn^2+ + Cu}\tag{R3}$$

$E_\mathrm{cell}^\circ = \pu{1.10 V}$ at $\pu{25 °C}$ and $\pu{1 atm}$ and $\pu{1 M}$ of solutions.

From the Nernst equation:

$$E = E_\mathrm{cell}^\circ - \frac{0.0591}{n}\ln Q\tag{1}$$

where $n$ is number of electrons and $Q$ is reaction quotient which is the concentrations of product solutions raised to the power of their coefficients over concentrations of reactant solutions raised to the power of their coefficients:

$$Q = \frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}\tag{2}$$

If you halve both the concentrations of $\ce{Zn^2+}$ and $\ce{Cu^2+}$, then $Q$ will remain the same and so will the the cell potential.


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