Solution

1. A Metallic tin is oxidized to (ce{Sn^{2+}(aq)}), and (ce{Be^{2+}(aq)}) is reduced to elemental beryllium. We can find the standard electrode potentials for the latter (reduction) half-reaction (−1.99 V) and for the former (oxidation) half-reaction (−0.14 V) directly from Table (PageIndex{1}).

B Adding the two half-reactions gives the overall reaction: [begin{align*}textrm{cathode} &: ce{Be^{2+}(aq)} +ce{2e^-} rightarrow ce{Be(s)} textrm{anode} &: ce{Sn(s)} rightarrow ce{Sn^{2+}(s)} +ce{2e^-} hline textrm{overall} &: ce{Sn(s)} + ce{Be^{2+}(aq)} rightarrow ce{Sn^{2+}(aq)} + ce{Be(s)} end{align*} ] with [begin{align*} E^circ_{textrm{cathode}} &=textrm{-1.99 V} [4pt] E^circ_{textrm{anode}} &=textrm{-0.14 V} [4pt] E^circ_{textrm{cell}} &=E^circ_{textrm{cathode}}-E^circ_{textrm{anode}} [4pt] &=-textrm{1.85 V} end{align*} ]

The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot reduce Be2+ to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn2+) by metallic beryllium, which has a positive value of E°cell, will occur spontaneously.

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1. A (ce{MnO2}) is the oxidant ((ce{Mn^{4+}}) is reduced to (ce{Mn^{2+}})), while (ce{H2O2}) is the reductant ((ce{O^{2−}}) is oxidized to (ce{O2})). We can obtain the standard electrode potentials for the reduction and oxidation half-reactions directly from Table (PageIndex{1}).

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B The two half-reactions and their corresponding potentials are as follows: [begin{align*}textrm{cathode} &: ce{MnO_2(s)}+ce{4H^+(aq)}+ce{2e^-} rightarrowce{Mn^{2+}(aq)}+ce{2H_2O(l)} ce{anode} &: ce{H_2O_2(aq)}rightarrowce{O_2(g)}+ce{2H^+(aq)}+ce{2e^-} hline textrm{overall} &: ce{MnO_2(s)}+ce{H_2O_2(aq)}+ce{2H^+(aq)}rightarrowce{O_2(g)}+ce{Mn^{2+}(aq)}+ce{2H_2O(l)} end{align*} ] with [begin{align*} E^circ_{textrm{cathode}} &=textrm{1.23 V} [4pt] E^circ_{textrm{anode}} &=textrm{0.70 V} [4pt] E^circ_{textrm{cell}} &=E^circ_{textrm{cathode}}-E^circ_{textrm{anode}} [4pt] &=+textrm{0.52 V} end{align*} ]

The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce (ce{MnO2}), and oxygen gas will evolve from the solution.

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