Question

Sketch a cell that forms iron metal from iron(II) while changing chromium metal to chromium(III). Calculate the voltage, show the electron flow, label the anode and cathode, and balance the overall cell equation.

Short answer

The cell in which an iron metal formed from iron (III) and chromium metal to chromium (III) has been drawn. The voltage of cell can be calculated. The electron flow and labelling of cathode and anode needs to be shown. The overall cell equal need to be balanced.

Step-by-step solution

Definition of galvanic cell

The electrochemical cell that has transfer from electrical to chemical reactions within a cell. There is a separation of oxidation and reduction cells by wire connection and electrons flow through the wire.

Reduction half reaction of iron and chromium

The reduction half of iron equation can be written as,

$F{e}^{2+}+2e^{-}\to Fe{E}^o=-0.44$

The reduction half of chromium equation can be written as,

$C{r}^{3+}+3e^{-}\to Cr{E}^o=-0.73$

Chromium has less E^o value than Iron. The chromium can be oxidized and reduced.

Cell structure that forms iron metal from iron (II) to chromium (III)

The cell structure that forms iron metal from iron (II) to chromium (III) can be drawn as,

The diagram shows the electron flow from Chromium to Iron

The equation can be written as,

$\begin{array}{c}3F{e}^{2+}+6e^{-}\to 3Fe{E}^o=-0.44\\ Cr\to 2C{r}^{3+}+6e^{-}{E}^o=0.73\end{array}$

The overall balanced equation can be written as,

$2Cr+3F{e}^{2+}\to 2C{r}^{3+}+Fe{E^o}_{cell}=0.29V$