1876 United States presidential election in Iowa

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United States presidential election in Iowa, 1876

← 1872 November 7, 1876 1880 →
Turnout24.57% Increase 6.45 pp[1]
  RutherfordBHayes.png SamuelJonesTilden.png
Nominee Rutherford B. Hayes Samuel J. Tilden
Party Republican Democratic
Home state Ohio New York
Running mate William A. Wheeler Thomas A. Hendricks
Electoral vote 11 0
Popular vote 171,326 22,927
Percentage 58.50% 38.28%

President before election

Ulysses S. Grant

Elected President

Rutherford B. Hayes

The 1876 United States presidential election in Iowa took place on November 7, 1876, as part of the 1876 United States presidential election. Iowa voters chose eleven representatives, or electors, to the Electoral College, who voted for president and vice president.[2]

Iowa was won by Rutherford B. Hayes, the governor of Ohio (R-Ohio), running with Representative William A. Wheeler, with 58.50% of the vote, against Samuel J. Tilden, the former governor of New York (DNew York), running with Thomas A. Hendricks, the governor of Indiana and future vice president, with 38.28% of the popular vote.[2]

The Greenback Party chose industrialist Peter Cooper and former representative Samuel Fenton Cary, received 3.22% of the vote. Iowa gave the Greenbacks both their largest number of total votes and percentage and Iowa would later be the last bastion of the congressional Greenbacks who held a House seat until 1889.


United States presidential election in Iowa, 1876[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Rutherford B. Hayes of Ohio William A. Wheeler of Indiana 171,326 58.50% 11 100.00%
Democratic Samuel J. Tilden of New York Thomas A. Hendricks of Indiana 112,121 38.28% 0 0.00%
Greenback Peter Cooper of New York Samuel Fenton Cary of Ohio 9,431 3.22% 0 0.00%
Total 292,878 100.00% 11 100.00%


  1. ^ "1876 Presidential Election Results Iowa turnout".
  2. ^ a b c "1876 Presidential Election Results Iowa".