Question

Which of the following reactions is possible at anode?
A.) $2C{r^{3 + }} + 7{H_2}O \to C{r_2}O_7^{2 - } + 14{H^ + }$
B.) ${F_2} \to 2{F^ - }$
C.) $\dfrac{1}{2}{O_2} + 2{H^ + } \to {H_2}O$
D.) None of these.

Answer 1

The oxidation reaction occurs at anode terminal of the cell. The oxidation reaction takes place when electrons are lost and when the oxidation number of the elements increases in the reaction.

Complete step by step answer:
In the question, anode means the electrode of a battery that releases electrons that is at anode there is loss of electrons.
The oxidation is the process of losing electrons or gain of oxygen or loss of hydrogen takes place in it. The oxidation number of atoms increases in the oxidation process.
Opposite to it there is a cathode which absorbs electrons that is gain of electrons takes place in it.
Also, reduction is the process of gain of electrons or gain of hydrogen or also we can say loss of oxygen. The oxidation number decreases in this process.
So, we can say that:
At anode – Oxidation – Lose of electrons – Oxidation number increases.
At cathode – Reduction – Gain of electrons – Oxidation number decreases.
In option A.)
$2C{r^{3 + }} + 7{H_2}O \to C{r_2}O_7^{2 - } + 14{H^ + }$
Here as we can see at left side $Cr$ has oxidation number $+ 3$ and on right side the oxidation number of $Cr$ is $+ 6$.
Oxidation number of chromium is $+ 6$ in $C{r_2}O_7^{2 - }$ because there are two chromium atoms and seven oxygen atoms having $- 2$ charge on each and total charge is $- 2$. So, we can say that the sum of all charges in it is $- 2$. Let the oxidation number of chromium be $x$. Now,
$2x + 7( - 2) = - 2 \\\ 2x = - 2 + 14 \\\ 2x = 12 \\\ x = + 6 \\\$
Hence, we can say that the oxidation number of $Cr$ increases from $+ 3$ to $+ 6$ so oxidation takes place in this reaction.
For option B.)
${F_2} \to 2{F^ - }$
Here at left side fluorine has $0$ oxidation number and at right side fluorine has $- 1$ oxidation number. As oxidation number is decreasing thus reduction takes place in this reaction.
For option C.)
$\dfrac{1}{2}{O_2} + 2{H^ + } \to {H_2}O$
Here in this option, at left side oxygen has $0$ oxidation number and at right side it has $- 2$ oxidation number. As oxidation number is decreasing in this process hence, reduction occurs in this reaction.
Therefore the correct option is A.

Note:
Remember that in case of electrolytic cell, anode is the positive electrode and cathode is the negative electrode but in case of galvanic cell, anode is the negative electrode and cathode is the positive electrode but in both the cell oxidation occurs at anode.