1908 United States presidential election in Iowa

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.

Main article: 1908 United States presidential election

1908 United States presidential election in Iowa

November 3, 1908

Nominee William Howard Taft William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate James S. Sherman John Worth Kern Electoral vote 13 0 Popular vote 275,209 200,771 Percentage 55.62% 40.58%

President before election Theodore Roosevelt Republican Elected President William Howard Taft Republican

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]
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

1. "1908 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved 23 December 2013.