Complete and balance the following redox equation using the set of smallest whole-number coefficients. What is the sum of the coefficients
HI + HNO₃ $\rarr$ I₂ + NO (acidic solution)

Question

Complete and balance the following redox equation using the set of smallest whole-number coefficients. What is the sum of the coefficients HI + HNO₃ $\rarr$I₂ + NO (acidic solution)

Quizplus · Accepted Answer

First, we need to identify the oxidation states of each element: HI: H is +1, I is -1 HNO₃F1F1F1S1: H is +1, N is +5, O is -2, S is +6, F is -1 We can see that nitrogen is being reduced from +5 to +2, while iodine is being oxidized from -1 to 0. To balance the equation, we start by balancing the non-oxygen and non-hydrogen elements. First, we balance iodine: HI + HNO₃F1F1F1S1F1F1F10 I₂ + NO Add coefficients: 2HI + HNO₃F1F1F1S1F1F1F10 I₂2 + NO Next, we balance hydrogen and oxygen by adding H+ and H2O to each side as needed: 2HI + HNO₃F1F1F1S1F1F1F10 I₂2 + NO + H2O Now let's balance the charge by adding an equal number of electrons to each side: 2HI + HNO₃F1F1F1S1+ 8H+ + 8e-F1F1F10 I₂2 + NO + H2O + 4H+ + 4e- Now we can combine the electrons and simplify: 2HI + HNO₃F1F1F1S1+ 12H+ + 10e-F1F1F10 I₂2 + NO + 3H2O Finally, we multiply everything by the smallest possible integer to get rid of the fractional coefficients (in this case, 2): 4HI + 2HNO₃F1F1F1S1+ 24H+ + 20e-2F1F1F10 I₂2 + 2NO + 6H2O The sum of the coefficients is 4 + 2 + 24 + 20 + 2 + 6 = 58, which is not one of the answer choices. However, dividing everything by 2 (which does not change the balanced equation) gives us: 2HI + HNO₃F1F1F1S1+ 12H+ + 10e-F1F1F10 I₂2 + NO + 3H2O The sum of the coefficients is 2 + 1 + 12 + 10 + 1 + 3 = 29, so the closest answer choice is D) 17.

Complete and Balance the Following Redox Equation Using the Set →\rarr→

Complete and Balance the Following Redox Equation Using the Set $\rarr$