1876 United States presidential election in Iowa
| 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 | 112,121 |
| Percentage | 58.50% | 38.28% |
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Hayes
Tilden
Elected President
The 1876 United States presidential election in Iowa took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose 11 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 (D–New 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.
Results
[edit ]| 1876 United States presidential election in Iowa[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Rutherford B. Hayes of Ohio | William A. Wheeler of New York | 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% | ||||
See also
[edit ]References
[edit ]
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