# Does the concentration affect the voltage in electrochemistry

Hello for my final chemistry lab we are doing electrochemistry. For my independent lab I'm using 10mL of ZnSO4 and 10mL of CuSO4. Then connect them with a salt bridge. I connect Zn metal and Cu metal to a voltmeter and dip those metals in the correct beakers. The voltage I get is 1.044. However when I add 10mL of water to both beakers, I still receive the same result. Does diluting two solutions affect the voltage?

• Do you have reason not to trust your experimental results? – ericksonla Jun 1 '18 at 17:44
• What result were you expecting? What does the Nernst equation say? – 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:

$$Cu^{+2}+ 2e^{-} \longrightarrow Cu$$

$E_{1}= 0.34 V$

$$Zn^{+2} +2e^{-} \longrightarrow Zn$$

$E_{2}= -0.76 V$

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

Combining the half equations gives:

$$Zn + Cu^{+2} \longrightarrow Zn^{+2}+ Cu$$
$E_{cell}= 1.10V$ at $25^{o}C$ and $1atm$ and $1M$ of solutions

From the Nernst equation: $$E = E_{cell} - \frac{0.0591}{n} (ln(Q))$$

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.

$$Q=\frac{Zn^{+2}}{Cu^{+2}}$$

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