Cr in Cr2O7-2 has oxidation state of +6
Cr in Cr+3 has oxidation state of +3
So, Cr in Cr2O7-2 is reduced to Cr+3
I in I- has oxidation state of -1
I in IO3- has oxidation state of +5
So, I in I- is oxidised to IO3-
Answer: A
20. In the following reaction +IO 3(aq) 4H2O) 2Cr(aq) Cr2Oag +Iag) +8H (aq) is reduced. 7(aq)...
For the following reaction: 2KMnO4 (aq) + 8HBr (aq) –> 2MnO2 (aq) + 4H2O (l) +3Br2 (aq) + 2KBr (aq) a) what is being oxidized and what is being reduced? b) what is the oxidizing agent and what is the reducing agent? 3) give the oxidation number of each atom in the equation.
1. Consider the reaction below: 3Ag2S(s) + 8H+(aq) + 2NO3-(aq) ? 6Ag+(aq) + 3S(s) + 2NO(g) + 4H2O(l) In this reaction, which species is reduced? 2. Calculate Ecell for the following electrochemical cell which is operating under nonstandard conditions: Cr | Cr3+(0.010 M) || Ag+(0.00010 M) | Ag The relevant standard reduction potentials are: Cr3+(aq) + 3e- ? Cr(s) Eº = ?0.74 V Ag+(aq) + e- ? Ag(s) Eº = +0.80 V
Determine if Cr is oxidized, Given Problem. reduced, or neither. croy² (aq) + 2 H₂O + (aq) → Cr ₂ 0₂²-lag) | - + 3H2010 * The answer is neither oxidized or reduced I but I do not understand why. * +6 2- My reasoning :) +7 2- 2 croy2- → Cr2O72- If cr went from +7 to +6 it gained e- .: Cr i reduced. My math to determine charge : I 20* + 702- 2er* +80² 2x -...
Consider the following cell reaction: 2Cr(s) + 6H*(aq) - 2Cr3+ (aq) + 3H2(g); Eºcell = 0.74V Under standard-state conditions, what is Eº for the following half-reaction? C++ (aq) + 3e - Cr(s) Select one: a. -0.37 V Ob. 0.74 V Oc. -0.74 V Od 0.25V o e. 0.37 v
Please fast, test
Which species is reduced in the reaction below? Gamma(aq) + ClO^-(aq) rightarrow IO^- (aq) + Cl^- (aq) Gamma H_2O Cl^- IO^- ClO^- What is the oxidation number of each atom in sodium hydrogen carbonate, NaHCO_3? Na = +1, H = -1, C = + 6, O = -2 Na = +1, H = +1, C = +4, O = -2 Na = +1, H = -1, C = +2, O = -2 Na = 0, H =...
3. Consider the unbalanced reaction NO3- (aq) + Sn2+ (aq) +Snº+(aq) + NO(g) a. Atom In the following table, identify which atom is oxidized, which atom is reduced, and their oxidation states in the reactants and products. (12 points) Oxidation State in Oxidation State in Reactant Product Oxidized Reduced b. Balance the reaction under BASIC conditions. (15 points) NO3- (aq) + Sn2+ (aq) Sn"+ (aq) + NO(g)
help with these 3 please?
18. Of the reactions below, which one is a precipitation reaction? a. NH4Cl → NH3 + HCI b. 2Mg + O2 + 2MgO c. 2N2 + 3H2 + 2NH3 d. 2CH4 +402 2002 + 4H20 e. Ca(NO3)2 + Na2S → CSS(s) + 2NaNO3 19. In the equation, CrO42(aq) + H2O(l) → CrO2(aq) + OH-(ag), the change in the oxidation number of the chromium atom is a. decrease by six units b. decrease by three units...
Consider the following cell reaction at 18°C: Ca(e)+Cu+ (aq) + Ca2+ (aq) + Cu() Calculate the standard cell potential of this cell from the standard electrode potentials, and from this, obtain AG" for the cell reaction. Calculate AF. Use these values of AN and AG to obtain AS for the cell reaction. Ca²+ (aq) +20 + Ca() --2.76 V Cu? (g) +20 + Cu(s) - 0.84 V AH;(O.*()) -- 542.8 kJ/mol AH;(Out (as)) - 64.8 kJ/mol V AG- AH- kJ...
Consider the following UNBALANCED reaction IN ACIDIC SOLUTION: Fe2+(aq) + MnO4-(aq) ⟶ Fe3+(aq) + Mn2+(aq) a. (5) Species that is oxidized (be specific – i.e. identify which atom and if there are multiple atoms with different charges, identify the correct one) b. (5) Species that is reduced (same instructions as above) c. (10) Full Balanced oxidation ½ reaction (‘full’ means with regard to mass (atoms) and charges.) d. (10) Full Balanced reduction ½ reaction e. (9) Complete Balanced Reaction Using...
For the following redox reaction, Fe3O4(s) + H2(g) + 3 Fe(s) + 4H2O(1) match the chemical species on the right that fits the description on the left. The species oxidized in the reaction. [Choose ] The species reduced in the reaction. [Choose ] The oxidizing agent. [Choose ] The reducing agent. [Choose ] Identify the oxidation state of each element in ammonium chlorate, NH4C104. N (Choose ) H [Choose ] СІ [Choose) о [Choose ]