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(A) The given half-reaction is the reduction of the permanganate ion in an acidic medium.Step $1$: Balance the $Mn$ atoms.$2MnO_4^- + yH^+ + ze^- 	o

Vedclass · Accepted Answer

$2, 16, 10$

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(A) The given half-reaction is the reduction of the permanganate ion in an acidic medium.Step $1$: Balance the $Mn$ atoms.$2MnO_4^- + yH^+ + ze^- 	o 2Mn^{2+} + 4H_2O$Step $2$: Balance the oxygen atoms by adding $H_2O$.Since there are $8$ oxygen atoms on the left $(2 	imes 4)$,we need $8$ oxygen atoms on the right. The equation already has $4H_2O$,so we need $8H_2O$ on the right.$2MnO_4^- + yH^+ + ze^- 	o 2Mn^{2+} + 8H_2O$Step $3$: Balance the hydrogen atoms.There are $16$ hydrogen atoms on the right $(8 	imes 2)$,so $y = 16$.$2MnO_4^- + 16H^+ + ze^- 	o 2Mn^{2+} + 8H_2O$Step $4$: Balance the charge.Left side charge: $2(-1) + 16(+1) + z(-1) = -2 + 16 - z = 14 - z$Right side charge: $2(+2) + 0 = 4$$14 - z = 4 \implies z = 10$Thus,$x = 2, y = 16, z = 10$. Since the provided options do not match this result exactly,we identify the closest logical structure. Given the standard reduction half-reaction $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$,multiplying by $2$ gives $2MnO_4^- + 16H^+ + 10e^- 	o 2Mn^{2+} + 8H_2O$.

For the following balanced equation,$x, y$ and $z$ will be respectively:
$xMnO_4^- + yH^+ + ze^- \to 2Mn^{2+} + 4H_2O$

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For the following balanced equation,$x, y$ and $z$ will be respectively:$xMnO_4^- + yH^+ + ze^- \to 2Mn^{2+} + 4H_2O$