
Number of moles of ${K_2}C{r_2}{O_7}$ reduced by one mole of $S{n^{2 + }}$ ions is?
(A) $\frac{1}{3}$
(B) $3$
(C) $\frac{1}{6}$
(D) $6$
Answer
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Hint: In order to solve this question you must be aware of the concepts of Oxidation and Reduction. And also you must know the reduction reaction of ${K_2}C{r_2}{O_7}$ with $S{n^{2 + }}$ ions. Firstly, write the complete balanced net reaction using the two half reactions and then use the unitary method to find the number of moles of ${K_2}C{r_2}{O_7}$ reduced by one mole of $S{n^{2 + }}$ ions.
Complete step-by-step solution:
Step 1: In this step we will write the balanced chemical reactions (ionic reactions) for reduction of ${K_2}C{r_2}{O_7}$ and $S{n^{2 + }}$ :
\[C{r_2}{O_7}^{2 - } + \,\,14{H^ + }\, + \,\,6{e^ - } \to \,\,2C{r^{3 + }} + \,\,7{H_2}O\] ; this is the first half reaction
\[(S{n^{2 + }}\,\, \to \,\,S{n^{4 + }}\,\, + \,\,2{e^ - }) \times 3\] ; this is the second half reaction
And after combining these two reactions we can write the complete reaction.
Step 2: In this step we will write the Net balanced chemical reaction:
\[3S{n^{2 + }}\,\, + \,\,C{r_2}{O_7}^{2 - }\,\, + \,\,14{H^ + }\,\, \to \,\,3S{n^{4 + }}\,\, + \,\,2C{r^{3 + }}\,\, + \,\,7{H_2}O\]
Step 3: In this step we will use the unitary method to find the number of moles of ${K_2}C{r_2}{O_7}$ reduced by one mole of $S{n^{2 + }}$ ions:
According to the Net balanced reaction: 1 mole of $C{r_2}{O_7}^{2 - }$ will be reduced by 3 moles of $S{n^{2 + }}$
Therefore 1 mole of $S{n^{2 + }}$ will reduce = $\dfrac{1}{3}$ moles of $C{r_2}{O_7}^{2 - }$
Hence, $\dfrac{1}{3}$ is the required answer.
Therefore, the correct answer is option A.
Note:Redox reactions are characterized by the transfer of electrons between chemical species, most often with one species undergoing oxidation while another species undergoes reduction. Oxidation is the loss of electrons or an increase in the oxidation state of an atom, an ion, or of certain atoms in a molecule. Reduction is the gain of electrons or a decrease in the oxidation state of an atom, an ion, or of certain atoms in a molecule.
Complete step-by-step solution:
Step 1: In this step we will write the balanced chemical reactions (ionic reactions) for reduction of ${K_2}C{r_2}{O_7}$ and $S{n^{2 + }}$ :
\[C{r_2}{O_7}^{2 - } + \,\,14{H^ + }\, + \,\,6{e^ - } \to \,\,2C{r^{3 + }} + \,\,7{H_2}O\] ; this is the first half reaction
\[(S{n^{2 + }}\,\, \to \,\,S{n^{4 + }}\,\, + \,\,2{e^ - }) \times 3\] ; this is the second half reaction
And after combining these two reactions we can write the complete reaction.
Step 2: In this step we will write the Net balanced chemical reaction:
\[3S{n^{2 + }}\,\, + \,\,C{r_2}{O_7}^{2 - }\,\, + \,\,14{H^ + }\,\, \to \,\,3S{n^{4 + }}\,\, + \,\,2C{r^{3 + }}\,\, + \,\,7{H_2}O\]
Step 3: In this step we will use the unitary method to find the number of moles of ${K_2}C{r_2}{O_7}$ reduced by one mole of $S{n^{2 + }}$ ions:
According to the Net balanced reaction: 1 mole of $C{r_2}{O_7}^{2 - }$ will be reduced by 3 moles of $S{n^{2 + }}$
Therefore 1 mole of $S{n^{2 + }}$ will reduce = $\dfrac{1}{3}$ moles of $C{r_2}{O_7}^{2 - }$
Hence, $\dfrac{1}{3}$ is the required answer.
Therefore, the correct answer is option A.
Note:Redox reactions are characterized by the transfer of electrons between chemical species, most often with one species undergoing oxidation while another species undergoes reduction. Oxidation is the loss of electrons or an increase in the oxidation state of an atom, an ion, or of certain atoms in a molecule. Reduction is the gain of electrons or a decrease in the oxidation state of an atom, an ion, or of certain atoms in a molecule.
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