This half-reaction I2 + 2e− → 2I− represents the process of ____ because electrons are____ .
This half-reaction I2 + 2e− → 2I− represents the process of ____ because electrons are____ .
Half-reaction E° (V) I2(s) + 2e- 2I-(aq) 0.535V Pb2+(aq) + 2e- Pb(s) -0.126V Cr3+(aq) + 3e- Cr(s) -0.740V The strongest oxidizing agent is: ______enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will I2(s) reduce Cr3+(aq) to Cr(s)? Which species can be reduced by Pb(s)? If none, leave box blank.
If [I-] = 0.050 M, calculate the half-cell potential for the following reaction: I2 (s) + 2e- ↔ 2I- E° = +0.540 V
Use the standard half-cell potentials given, calculate the initial cell potential for the following reaction occurring in an electrochemical cell at 25°C when the starting concentrations of [Sn2+] = 2.0 M and [I2] = 0.010 M (The equation is balanced.) <Missing data [I−]=1.0M> I2 + Sn --> 2I- + Sn^2+ I2 + 2e- --> 2I- E=+0.54V Sn^2+ +2e- --> Sn E= -0.14V A) +0.47V B) +0.068V C) +0.75V D) +0.61V
Based on the half reactions given below, what is the E°cell, in volts, for the reaction performed in Experiment 22 (2Fe3+(aq) + 2I-(aq) ⟶ 2Fe2+(aq) + I2(aq))? Enter your answer to the thousandths place. Fe3+(aq) + e- ⟶ Fe2+(aq) E°red = 0.771 V I2(aq) + 2e- ⟶ 2I-(aq) E°red = 0.621 V
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq) 1.36 O2(g)+4H+(aq)+4e? ?2H2O(l) 1.23 Br2(l)+2e? ?2Br?(aq) 1.09 NO3?(aq)+4H+(aq)+3e? ?NO(g)+2H2O(l) 0.96 Ag+(aq)+e? ?Ag(s) 0.80 I2(s)+2e? ?2I?(aq) 0.54 Cu2+(aq)+2e? ?Cu(s) 0.16 2H+(aq)+2e? ?H2(g) 0 Cr3+(aq)+3e? ?Cr(s) -0.73 2H2O(l)+2e? ?H2(g)+2OH?(aq) -0.83 Mn2+(aq)+2e? ?Mn(s) -1.18 How can the table be used to predict whether or not a metal will dissolve in HCl? In HNO3? Drag the terms on the left to the appropriate blanks on the right to...
For the reaction H2O2 + 2H+ + 2I – ® I2 + 2H2O the following mechanism has been suggested. H2O2 + I – ® HOI + OH – slow OH – + H+ ® H2O fast HOI + H+ + I – ® I2 + H2O fast Identify the rate law for the reaction and explain why
Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction: I2(g)⇌2I(g), Kc=0.011 at 1200∘C In an equilibrium mixture, the concentration of I2 is 0.10 M. What is the equilibrium concentration of I?
Identify the molecularity and the rate law for each of the following elementary reactions. (a) I2(g)→2I(g)I2(g)→2I(g) (b) 2NO(g)+Br2(g)→2NOBr(g)2NO(g)+Br2(g)→2NOBr(g) (c) N2O5(g)→NO2(g)+NO3(g)N2O5(g)→NO2(g)+NO3(g) Match the items in the left column to the appropriate blanks in the sentences on the right. Reaction (a) is , Rate = Reaction (b) is , Rate = reaction (c) is , rate = .
Be sure to answer all parts. The equilibrium constant Kc for the reaction I2(g) ⇆ 2I(g) is 3.84 × 10−5 at 724 ° C. Calculate Kc and KP for the equilibrium 2I(g) ⇆ I2(g) at the same temperature. Kc = × 10 (Enter your answer in scientific notation.) KP =
Consider the following balanced redox reaction. 2S2O32- + I2 → 2I- + S4O62- How many moles of I2 can be consumed by 1.45 mL of 0.15 M Na2S2O3?