# Question #b0597

Apr 24, 2015

Statement number (2) is correct, ${H}^{+}$ can oxidize iron and nickel can reduce $B {r}_{2}$.

Standard electrode potentials are used to express the ability of a substance to lose electrons.

When you compare the ${E}^{0}$ values of two species, keep in mind that a more positive (or less negative) ${E}^{0}$ will correspond to an oxidizer, and a more negative (or less positive) ${E}^{0}$ will correspond to a reducer.

Here's a link to a table of Standard Electrode Potentials: http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html

Any species that appears on the left of the reduction half-reaction will oxidize any species that appears on the right of a reduction half-reaction listed above it in the table.

Likewise, any species that appears on the right of the reduction half-reaction will reduce any species that appears on the left of the reduction half-reaction listed below it in the table.

SIDE NOTE This is true when negative values for ${E}^{0}$ are listed at the top of the table.

So, starting with the first reaction

• $B {r}_{2}$ can oxidize $N i$ and ${H}_{2}$ can reduce $M {n}^{2 +}$

The reduction half-reactions are

$B {r}_{2 \left(l\right)} + 2 {e}^{-} \to 2 B {r}_{\left(a q\right)}^{-}$, $\text{ "E^0 "= +1.07 V}$

$N {i}_{\left(a q\right)}^{2 +} + 2 {e}^{-} \to N {i}_{\left(s\right)}$, $\text{ "E^0 = "-0.23 V}$

Bromine can oxidize nickel because its reduction half-reaction is lower in the table, i.e. it has a more positive ${E}^{0}$ (+1.07 V compared with -0.23 V).

$2 {H}_{\left(a q\right)}^{+} + 2 {e}^{-} \to {H}_{2 \left(g\right)}$, $\text{ "E^0 = "0 V}$

$M {n}_{\left(a q\right)}^{2 +} + 2 {e}^{-} \to M {n}_{\left(s\right)}$, $\text{ "E^0 = "-1.18 V}$

Hydrogen gas cannot reduce $M {n}^{2 +}$ because the latter is located above it in the table, i.e. has a more negative ${E}^{0}$ (-1.18 V compared with 0V).

• ${H}^{+}$ can oxidize $F e$ and $N i$ can reducce $B {r}_{2}$

$2 {H}_{\left(a q\right)}^{+} + 2 {e}^{-} \to {H}_{2 \left(g\right)}$, $\text{ "E^0 = "0 V}$

$F {e}_{\left(a q\right)}^{2 +} + 2 {e}^{-} \to F {e}_{\left(s\right)}$, $\text{ "E^0 = "-0.41 V}$

${H}^{+}$ can oxidize anything on the right of the reduction half-reaction above it in the table, i.e. that has a more negative ${E}^{0}$.

As a result, ${H}^{+}$ can oxidize iron.

The last one will be obvious, since we've already established that bromine can oxidize nickel.

Nickel can reduce bromine because it has a more negative ${E}^{0}$ value, since it's located above it in the table.