
What is the reaction at anode during the electrolysis of copper [II] sulphate solution using platinum as cathode and carbon as anode?
A) \[C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)\]
B) \[Cu(aq) \to Cu(s) + 2{e^ - }\]
C) \[4OH(aq) + 4{e^ - } \to 2{H_2}0(l) + {O_2}(g)\]
D) \[4O{H^ - }(aq) \to 2{H_2}0(l) + {O_2}(g) + 4{e^ - }\]
Answer
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Hint: In electrochemistry, the anode is where oxidation occurs and is the positive polarity contact in an electrolytic cell. At the anode, anions (negative ions) when subjected to the electrical potential are forced to react chemically and eject electrons (oxidation) which then flow up and into the circuit of the experiment.
Complete step by step answer:
What mainly happens in electrolysis is the interchange of atoms and ions between the different components of the reaction. This process happens by the removal or addition of electrons due to the applied current. The output of the electrolysis product is the desired product which is usually in a different physical state as the electrolyte and can be removed by physical processes (e.g. we can collect the gas above an electrode or by precipitating a product out of the electrolyte).
Each electrode attracts ions that are of the opposite charge. Positively charged ions (cations) move towards the electron-providing (negative) cathode. Negatively charged ions (anions) move towards the electron-extracting (positive) anode. In this process electrons are effectively introduced at the cathode as a reactant and removed at the anode as a product. In chemistry, the loss of electrons is called oxidation, while electron gain is called reduction.
During the electrolysis of copper [II] sulphate solution dissociation of electrolyte happens as follows,
\[
CuS{O_4} \to C{u^{2 + }} + SO_4^{2 - } \\
{H_2}0 \to {H^ + } + O{H^ - } \\
\]
Reaction at cathode happens in the following manner:
\[C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)\]
Reaction at anode happens like this:
\[
O{H^ - } - 1{e^ - } \to OH \\
4O{H^ - }(aq) \to 2{H_2}0(l) + {O_2}(g) + 4{e^ - } \\
\]
Hence, the correct option is D.
Note:
Electrolysis (of copper sulphate) is a way of splitting up (decomposition) of the compound (copper sulphate) using electrical energy. The electrolysis of copper [II] sulphate can take place in presence of inert electrodes, copper electrodes or platinum electrodes. The outcome of these electrolysis is slightly different from each other and should be considered while solving such questions.
Complete step by step answer:
What mainly happens in electrolysis is the interchange of atoms and ions between the different components of the reaction. This process happens by the removal or addition of electrons due to the applied current. The output of the electrolysis product is the desired product which is usually in a different physical state as the electrolyte and can be removed by physical processes (e.g. we can collect the gas above an electrode or by precipitating a product out of the electrolyte).
Each electrode attracts ions that are of the opposite charge. Positively charged ions (cations) move towards the electron-providing (negative) cathode. Negatively charged ions (anions) move towards the electron-extracting (positive) anode. In this process electrons are effectively introduced at the cathode as a reactant and removed at the anode as a product. In chemistry, the loss of electrons is called oxidation, while electron gain is called reduction.
During the electrolysis of copper [II] sulphate solution dissociation of electrolyte happens as follows,
\[
CuS{O_4} \to C{u^{2 + }} + SO_4^{2 - } \\
{H_2}0 \to {H^ + } + O{H^ - } \\
\]
Reaction at cathode happens in the following manner:
\[C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)\]
Reaction at anode happens like this:
\[
O{H^ - } - 1{e^ - } \to OH \\
4O{H^ - }(aq) \to 2{H_2}0(l) + {O_2}(g) + 4{e^ - } \\
\]
Hence, the correct option is D.
Note:
Electrolysis (of copper sulphate) is a way of splitting up (decomposition) of the compound (copper sulphate) using electrical energy. The electrolysis of copper [II] sulphate can take place in presence of inert electrodes, copper electrodes or platinum electrodes. The outcome of these electrolysis is slightly different from each other and should be considered while solving such questions.
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