Question

when the following equation is balanced in an acidic solution how many electrons are transferred? enter only a numeric answer. for full marks in the dropbox you must show a final balanced redox reaction. x₂o₇²⁻ + y²⁺ → x³⁺ + y³⁺ note the following: o=oxygen x=metal y=metal

Explanation:

Step1: Split into half - reactions

Oxidation half - reaction: $Y^{2+}\to Y^{3+}+e^-$
Reduction half - reaction: $X_2O_7^{2 - }\to2X^{3+}$

Step2: Balance oxygen in reduction half - reaction

Add $7H_2O$ to the right side: $X_2O_7^{2 - }\to2X^{3+}+7H_2O$

Step3: Balance hydrogen in reduction half - reaction

Add $14H^+$ to the left side: $X_2O_7^{2 - } + 14H^+\to2X^{3+}+7H_2O$

Step4: Balance charge in reduction half - reaction

$X_2O_7^{2 - }+14H^++6e^-\to2X^{3+}+7H_2O$

Step5: Adjust oxidation half - reaction

Multiply the oxidation half - reaction by 6: $6Y^{2+}\to6Y^{3+}+6e^-$

Step6: Combine half - reactions

$X_2O_7^{2 - }+14H^++6Y^{2+}\to2X^{3+}+6Y^{3+}+7H_2O$

Answer:

6