Meta–chlorobenzaldehyde will undergo Cannizzaro reaction with $50\% \mathrm{KOH}$ to give mchlorobenzoate ion and $\mathrm{m}-$chlorobenzyl alcohol. 

Since $\mathrm{m}-$chlorobenzaldehyde does not contain α hydrogen thus with $50\% \mathrm{KOH}$ it undergoes the Cannizzaro reaction. The Cannizzaro reaction for the $\mathrm{m}-$chlorobenzaldehyde is given as:

Since it is a disproportionation reaction so one molecule of $\mathrm{m}-$chlorobenzaldehyde is reduced to $\mathrm{m}-$chlorobenzyl alcohol at the cost of others which is oxidized potassium $\mathrm{m}-$chlorobenzoate.