Question

How can we represent the half cell involving the following reaction?
$C{r_2}{O_7}^{2 - }(aq) + 14{H^ + }(aq) + 6{e^-} \to 2C{r^{3 + }}(aq) + 7{H_2}O$
(A) $Pt,{H^ + }(aq),C{r_2}{O_7}^{2 - }(aq)$
(B) $Pt,{H^ + },C{r^{3 + }}(aq)$
(C) $Pt,{H_2},C{r_2}{O_7}^{2 - }(aq),C{r^{ + 3}}(aq)$
(D) $Pt,{H^ + },C{r_2}{O_7}^{2 - }(aq),C{r^{3 + }}(aq)$

Answer 1

Calculate the oxidation states of the central atom i.e. chromium before and after the reaction. With this information we can find whether the reaction is an oxidizing reaction or reducing reaction. While representing a reaction, the anode is placed on the left and cathode on the right.

Complete step-by-step answer: n the given reaction, we first calculate the oxidation state of Chromium on the reactants side and represent it as $X$ . The oxidation state of Oxygen is two.
Hence, $2X + 7( - 2) = - 2$
By solving this, we get the value of $X$ as $+ 6$
Similarly, we calculate the oxidation state of Chromium after the reaction and its value is $+ 3$
Since the oxidation state of chromium is reducing, the reaction is a reduction reaction. In electrochemical cell notation, the anode and cathode cells are separated by two bars or slashes and it represents a salt bridge. The anode is placed on the left and the cathode is placed on the right.
The above reaction is a reduction reaction and it occurs at the cathode side of the cell and hence we represent this reaction on the right side.

Therefore we can write the representation for this half cell as $Pt,{H^ + },C{r_2}{O_7}^{2 - }(aq),C{r^{3 + }}(aq)$ i.e. option (D).

Note: For a cell notation, the anode half-cell is described first and then the cathode half-cell. Usually oxidation reaction occurs at anode and reduction at the cathode. Therefore, the oxidation reaction is described first followed by the reduction reaction.