1908 United States presidential election in Iowa
Appearance
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Elections in Iowa |
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The 1908 United States presidential election in Iowa took place on November 3, 1908. Voters chose 13 representatives, or electors to the Electoral College, who voted for president and vice president.
Iowa voted for the Republican nominee, Secretary of War William Howard Taft, over the Democratic nominee, former U.S. Representative William Jennings Bryan. Taft won the state by a margin of 15.04%.
Results
1908 United States presidential election in Iowa[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | William Howard Taft of Ohio | James Schoolcraft Sherman of New York | 275,209 | 55.62% | 13 | 100.00% | ||
Democratic | William Jennings Bryan of Nebraska | John Worth Kern of Indiana | 200,771 | 40.582% | 0 | 0.00% | ||
Prohibition | Eugene Wilder Chafin of Illinois | Aaron Sherman Watkins of Ohio | 9,837 | 1.99% | 0 | 0.00% | ||
Socialist | Eugene Victor Debs of Indiana | Benjamin Hanford of New York | 8,287 | 1.67% | 0 | 0.00% | ||
Independence | Thomas Louis Hisgen of Massachusetts | John Temple Graves of Georgia | 404 | 0.08% | 0 | 0.00% | ||
Populist | Thomas Edward Watson of Georgia | Samuel Wardell Williams of Indiana | 261 | 0.05% | 0 | 0.00% | ||
Total | 494,769 | 100.00% | 13 | 100.00% |
References
- ^ "1908 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved 23 December 2013.